## Right Triangles

Rules, Formula and more

## Pythagorean Theorem

The sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse .

Usually, this theorem is expressed as $$A^2 + B^2 = C^2$$ .

## Right Triangle Properties

A right triangle has one $$90^{\circ}$$ angle ($$\angle$$ B in the picture on the left) and a variety of often-studied formulas such as:

• The Pythagorean Theorem
• Trigonometry Ratios (SOHCAHTOA)
• Pythagorean Theorem vs Sohcahtoa (which to use)

SOHCAHTOA only applies to right triangles ( more here ) .

## A Right Triangle's Hypotenuse

The hypotenuse is the largest side in a right triangle and is always opposite the right angle.

In the triangle above, the hypotenuse is the side AB which is opposite the right angle, $$\angle C$$.

Online tool calculates the hypotenuse (or a leg) using the Pythagorean theorem.

## Practice Problems

Below are several practice problems involving the Pythagorean theorem, you can also get more detailed lesson on how to use the Pythagorean theorem here .

Find the length of side t in the triangle on the left.

Substitute the two known sides into the Pythagorean theorem's formula : A² + B² = C²

What is the value of x in the picture on the left?

Set up the Pythagorean Theorem : 14 2 + 48 2 = x 2 2,500 = X 2

$$x = \sqrt{2500} = 50$$

$$x^2 = 21^2 + 72^2 \\ x^2= 5625 \\ x = \sqrt{5625} \\ x =75$$

Find the length of side X in the triangle on on the left?

Substitue the two known sides into the pythagorean theorem's formula : $$A^2 + B^2 = C^2 \\ 8^2 + 6^2 = x^2 \\ x = \sqrt{100}=10$$

What is x in the triangle on the left?

x 2 + 4 2 = 5 2 x 2 + 16 = 25 x 2 = 25 - 16 = 9 x = 3

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## Right triangle

Here you will learn about right triangles, including what a right triangle is and how to solve problems with right triangles.

Students first learn about right triangles in the 4th grade with their work in geometry.

## What is a right triangle?

A right triangle is a type of triangle that has one right angle ( 90^{\circ} angle).

The two sides of the triangle that meet at the right angle are perpendicular. In this case, side AB is perpendicular to side BC so angle B is 90^{\circ}.

A right angle is 90^{\circ} and it is symbolized using a small square inside the angle at the vertex.

The longest side of the right triangle is called the hypotenuse . It is the side opposite to the right angle.

As you move into secondary school and study geometry, triangles are one of the most common shapes to recognize for angles in parallel lines , circle theorems , interior angles , trigonometry , Pythagoras’ theorem and many more.

## Types of right triangles

There are two types of right triangles.

## Common Core State Standards

How does this apply to 4th and 5th grade math?

• Grade 4 – Geometry (4.G.A.1) Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures.
• Grade 4 – Geometry (4.G.A.2) Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles.
• Grade 5 – Geometry (5.G.B.3) Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category.

## How to solve problems involving right angle triangles

In order to identify right triangles:

Recall the definition.

Explain how the triangle fits or does not fit the definition.

## [FREE] Triangles Check for Understanding Quiz (Grade 4 to 5)

Use this quiz to check your grade 4 to 5 students’ understanding of triangles. 10+ questions with answers covering a range of 4th and 5th grade triangle topics to identify areas of strength and support!

## Right triangle examples

Example 1: identifying right triangles.

Is the triangle a right triangle?

A right triangle has one right angle (90^{\circ}).

2 Explain how the triangle fits or does not fit the definition.

There is one angle that appears to be 90^{\circ}. Measuring the angle on the protractor, it measures to be 90^{\circ}. So, the triangle is a right triangle.

## Example 2: identifying right triangles

There does not appear to be any angle that looks 90^{\circ}. The angle looks obtuse and measures to be 110^{\circ}. The other two angles are acute. So the triangle is not a right triangle.

## Example 3: identifying right triangles

Is triangle ABC a right triangle if side AB is perpendicular to side BC?

Perpendicular lines intersect (meet) to form a right angle.

Since AB and BC are perpendicular, a right angle (90^{\circ}) is formed at B.

Triangle ABC is a right triangle.

## Example 4: identifying right triangles

Ana drew a triangle where two of the sides of the triangle are perpendicular. Is the triangle Ana drew a right triangle?

The triangle Ana drew is a right triangle because there is a right angle (90^{\circ}).

## Example 5: identifying right triangles

Is the triangle below a right scalene triangle or a right isosceles triangle?

A scalene right triangle is a right triangle where the sides of a triangle have no equal sides and no equal angles.

The triangle has one right angle and three sides that have different measures. So, the right triangle is a right scalene triangle.

## Example 6: identifying right triangles

A right isosceles triangle is a right triangle with two equal sides and two equal angles.

The triangle has a right angle with two sides that are marked as equal and two angles that are marked as equal, so the triangle is a right isosceles triangle.

## Teaching tips for right triangles

• Have students use manipulatives to investigate triangle properties.
• Facilitate learning opportunities like projects where students can see how right triangles are used in the real world.
• Integrate drawing activities where students can use rulers and protractors.

## Easy mistakes to make

• Confusing angle facts Make sure to recall that an acute angle is an angle greater than 0^{\circ} and less than 90^{\circ}; obtuse angle is an angle 90^{\circ} and less than 180^{\circ}; and a right angle is equal to 90^{\circ}.
• Thinking that if an angle looks like it’s \bf{90°} then it is a right angle In geometry, the angle has to be marked as 90^{\circ} or you have to measure it to be 90^{\circ} to be certain that it is a right angle.

## Related triangles lessons

• Types of triangles
• Acute triangles (coming soon)
• Equilateral triangle
• Isosceles triangles
• Scalene triangles

## Practice right triangle questions

1. Select the right triangle.

A right triangle is a triangle with a right angle. This triangle appears to have a right angle.

Measuring the angle with a protractor, you can see that it is 90^{\circ} which means the triangle is a right triangle.

2. Select the right isosceles triangle.

Right isosceles triangles have one right angle, two equal sides, and 2 equal angles.

This triangle has a right angle, two angles that measure 45^{\circ} and two sides that measure 5 inches.

3. Which triangle has a set of perpendicular sides?

Perpendicular lines are lines that intersect (meet) at a right angle (90^{\circ}).

This triangle has two sides that are perpendicular because there is a square symbol that means 90^{\circ} at the vertex.

4. Which photo is a right triangle?

A right triangle has 1 right angle.

This triangle has three angles that are all acute, less than 90^{\circ}.

This triangle has an angle that is obtuse, greater than 90^{\circ} but less than 180^{\circ}.

This triangle is the right triangle because it has one 90^{\circ} angle.

5. Nora drew a triangle that has a pair of perpendicular sides. Which triangle could be the one Nora drew?

This is the triangle that Nora could have drawn because it’s the only triangle that has a right angle.

Perpendicular lines intersect to form right angles.

6. Which triangle represents a right scalene triangle?

A right scalene triangle is a triangle with no equal sides, no equal angles, and 1 right angle.

This triangle is a right triangle but there are two sides of the triangle marked as equal.

This triangle is the right scalene triangle because the sides have different measurements and there is one right angle.

## Right triangle FAQs

The Pythagorean theorem is a formula that shows the relationship of the side lengths of right triangles, a^2+b^2=c^2 where the sum of (side a ) ^2 and (side b ) ^2 is equal to (hypotenuse c ) ^2. (The sum of the squares of the sides is equal to the square of the hypotenuse.) You will learn how to use the Pythagorean theorem in 8th grade.

Trigonometry is the study of triangles. In trigonometry, you learn more about right triangles and the ratio of the sides of a right triangle – sine, cosine, and tangent. You will also learn about trigonometric functions.

A Pythagorean triple is composed of three positive integers that work in the Pythagorean theorem. For example, the numbers 3, 4, 5 are considered a Pythagorean triple because the sum of the squares of the sides is equal to the square of the 3^2+4^2=5^2 → 9+16 =25.

Special right triangles will be a topic you explore in high school. There are two special right triangles. The first one is an isosceles right triangle or 45-45-90 triangle. The other one is a scalene right triangle, or 30-60-90 triangle.

The way you find the area of a right triangle is the same way you find the area of oblique triangles. You can use the formula for the area of the triangle, A=\cfrac{1}{2} \, \cdot b \cdot h.

Angles can be measured in two different units, degrees or radians. You will learn how to measure angles in radians in high school.

You find the perimeter of a right triangle by adding up the length of the sides and the length of the hypotenuse, which is the largest side of the triangle.

Only right triangles have a side called the hypotenuse. The hypotenuse of a right triangle is the longest side of the triangle.

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## Privacy Overview

• Trigonometry
• Study Guides
• Solving Right Triangles
• Functions of Acute Angles
• Functions of General Angles
• Tables of Trigonometric Functions
• Law of Cosines
• Law of Sines
• Solving General Triangles
• Areas of Triangles
• Circular Functions
• Periodic and Symmetric Functions
• Graphs: Sine and Cosine
• Graphs: Other Trigonometric Functions
• Graphs: Special Trigonometric Functions
• Introduction to Graphs
• Double‐Angle and Half‐Angle Identities
• Tangent Identities
• Product‐Sum and Sum‐Product Identities
• Fundamental Identities
• The Rectangular Coordinate System
• Vector Operations
• Geometry of Complex Numbers
• De Moivre's Theorem
• Polar Coordinates
• Inverse Cosine and Inverse Sine
• Other Inverse Trigonometric Functions
• Trigonometric Equations
• Simple Harmonic Motion
• The Expression M sin Bt + N cos Bt
• Uniform Circular Motion

Example 1 : Solve the right triangle shown in Figure (b) if ∠ B = 22°

Because the three angles of a triangle must add up to 180°, ∠ A = 90 ∠ B thus ∠ A = 68°.

The following is an alternate way to solve for sides a and c:

This alternate solution may be easier because no division is involved.

Example 2 : Solve the right triangle shown in Figure (b) if b = 8 and a = 13.

You can use the Pythagorean theorem to find the missing side, but trigonometric relationships are used instead. The two missing angle measurements will be found first and then the missing side.

Example 3: A large airplane (plane A) flying at 26,000 feet sights a smaller plane (plane B) traveling at an altitude of 24,000 feet. The angle of depression is 40°. What is the line of sight distance ( x ) between the two planes?

Figure 3 illustrates the conditions of this problem.

From Figure 3 , you can find the solution by using the sine of 40 °:

Example 4: A ladder must reach the top of a building. The base of the ladder will be 25′ from the base of the building. The angle of elevation from the base of the ladder to the top of the building is 64°. Find the height of the building (h) and the length of the ladder ( m ).

Figure 4 illustrates the conditions of this problem.

Example 5: A woodcutter wants to determine the height of a tall tree. He stands at some distance from the tree and determines that the angle of elevation to the top of the tree is 40°. He moves 30′ closer to the tree, and now the angle of elevation is 50°. If the woodcutter's eyes are 5′ above the ground, how tall is the tree?

From the small right triangle and from the large right triangle, the following relationships are evident:

Substituting the first equation in the second yields:

Note that 5′ must be added to the value of x to get the height of the tree, or 90.06′ tall.

Example 6: Using Figure 6, find the length of sides x and y and the area of the large triangle.

Because this is an isosceles triangle, and equal sides are opposite equal angles, the values of x and y are the same. If the triangle is divided into two right triangles, the base of each will be 6. Therefore,

Previous Areas of Triangles

Next Circular Functions

Chapter 2: Trigonometric Ratios

## 2.3 Solving Right Triangles

Algebra refresher.

1. $\:\sqrt{2}\sqrt{2}$

2. $\:\dfrac{3}{\sqrt{3}}$

3. $\:\sqrt{8}$

4. $\:\sqrt{\dfrac{3}{4}}$

Rationalize the denominator.

5. $\:\dfrac{1}{\sqrt{2}}$

6. $\:\dfrac{2}{\sqrt{3}}$

7. $\:\dfrac{6}{\sqrt{3}}$

8. $\:\dfrac{4}{\sqrt{8}}$

$\underline{\qquad\qquad\qquad\qquad}$

• $\displaystyle 2\vphantom{\dfrac{\sqrt{3}}{2}}$
• $\displaystyle \sqrt{3}\vphantom{\vphantom{\dfrac{\sqrt{3}}{2}}}$
• $\displaystyle 2\sqrt{2}\vphantom{\dfrac{\sqrt{3}}{2}}$
• $\displaystyle \dfrac{\sqrt{3}}{2}$
• $\displaystyle \dfrac{\sqrt{2}}{2}$
• $\displaystyle \dfrac{2\sqrt{3}}{3}$
• $\displaystyle 2\sqrt{3}\vphantom{\dfrac{\sqrt{3}}{2}}$
• $\displaystyle \sqrt{2}\vphantom{\dfrac{\sqrt{3}}{2}}$

Learning Objectives

• Solve a right triangle
• Use inverse trig ratio notation
• Use trig ratios to find an angle
• Solve problems involving right triangles
• Know the trig ratios for the special angles

## Introduction

A triangle has six parts: three sides and three angles. In a right triangle, we know that one of the angles is $90 {^o}\text{.}$ If we know three parts of a right triangle, including one of the sides, we can use trigonometry to find all the other unknown parts. This is called solving the triangle.

## Example 2.31.

The hypotenuse of a right triangle is 150 feet long, and one of the angles is $35 {^o}\text{,}$ as shown in the figure. Solve the triangle.

We can find the side opposite the 35° angle by using the sine ratio. \begin{align*} \sin 35 {^o} = \dfrac {\text{opposite}}{\text{hypotenuse}}\\ 0.5736 = \dfrac{a}{150}\\ a = 150(0.5736) = 86.04 \end{align*} The opposite side is about 86 feet long. To find side $b$, we could use the Pythagorean theorem now, but it is better to use given information, rather than values we have calculated, to find the other unknown parts. We will use the cosine ratio. \begin{align*} \cos 35 {^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}}\\ 0.8192 = \dfrac{b}{150}\\ b = 150(0.8192) = 122.89 \end{align*} The adjacent side is about $123$ feet long. Finally, the unknown angle is the complement of $35 {^o}\text{,}$ or $55 {^o}\text{.}$

## Checkpoint 2.32.

Sketch a right triangle with

• one angle of $37 {^o}\text{,}$ and
• the side adjacent to that angle of length 5 centimeters.

Without doing the calculations, list the steps you would use to solve the triangle.

Use tan $37{^o}$ to find the opposite side. Use cos $37{^o}$ to find the hypotenuse. Subtract $37{^o}$ from $90{^o}$ to find the third angle.

## Finding an Angle

While watching her niece at the playground, Francine wonders how steep the slide is. She happens to have a tape measure and her calculator with her and finds that the slide is 77 inches high and covers a horizontal distance of 136 inches, as shown below.

Francine knows that one way to describe the steepness of an incline is to calculate its slope, which in this case is $\begin{equation*} \dfrac {\Delta y}{\Delta x} = \dfrac{77}{136} = 0.5662 \end{equation*}$ However, Francine would really like to know what angle the slide makes with the horizontal. She realizes that the slope she has just calculated is also the tangent of the angle she wants.

If we know the tangent of an angle, can we find the angle? Yes, we can: locate the key labeled $\boxed{\text{TAN}^{-1}}$ on your calculator; it is probably the second function above the TAN key. Enter $\qquad\qquad\qquad$ 2nd TAN 0.5662

and you should find that $\begin{equation*} \text{tan}^{-1} 0.5662 = 29.52 {^o}\text{.} \end{equation*}$ This means that $29.52 {^o}$ is the angle whose tangent is $0.5662\text{.}$ We read the notation as “ inverse tangent of $0.5662$ is $29.52$ degrees.”

When we find $\tan^{-1}$ of a number, we are finding an angle whose tangent is that number. Similarly, $\sin^{-1}$ and $cos^{-1}$ are read as “inverse sine” and “inverse cosine.” They find an angle with the given sine or cosine.

Example 2.33.

Find the angle whose sine is $0.6834\text{.}$

Enter 2nd SIN 0.6834 into your calculator to find $\begin{equation*} \sin^{-1} 0.6834 = 43.11{^o} \end{equation*}$ So $43.11{^o}$ is the angle whose sine is $0.6834\text{.}$ Or we can say that $\begin{equation*} \sin 43.11{^o} = 0.6834 \end{equation*}$ You can check the last equation on your calculator.

In the last example, the two equations:

$\begin{equation*} \sin 43.11{^o} = 0.6834\end{equation*}$

$\begin{equation*} \sin^{-1} 0.6834 = 43.11{^o}\end{equation*}$

say the same thing in different ways.

## Caution 2.35.

The notation $\sin^{-1} x$ does not mean $\dfrac{1}{\sin x}\text{.}$ It is true that we use negative exponents to indicate reciprocals of numbers; for example, $a^{-1} = \dfrac{1}{a}$ and $3^{-1} = \dfrac{1}{3}\text{.}$ But “sin” by itself is not a variable.

• $\sin^{-1} x$ means “the angle whose sine is $x$”
• $\dfrac{1}{\sin x}$ means “the reciprocal of the sine of angle $x$”

(You may recall that $f^{-1}(x)$ denotes the inverse function for $f(x)\text{.}$ We will study trigonometric functions in Chapter 4.)

## Checkpoint 2.36.

Write the following fact in two different ways: $68{^o}$ is the angle whose cosine is $0.3746\text{.}$

$\cos 68{^o} = 0.3746$ or $\cos^{-1}(0.3746) = 68{^o}$

Example 2.37.

Find the angle of inclination of a hill if you gain $400$ feet in elevation while traveling half a mile.

A sketch of the hill is shown at right. (Recall that $1$ mile $= 5280$ feet.)

\begin{align*} \sin \theta = \dfrac {400}{2640} = 0.\overline{15}\\ \theta = \sin^{-1} 0.\overline{15} = 8.71{^o} \end{align*}

The angle of inclination of the hill is about $8.7{^o}\text{.}$

## Checkpoint 2.38.

The tallest living tree is a coastal redwood named Hyperion, at 378.1 feet tall. If you stand 100 feet from the base of the tree, what is the angle of elevation of your line of sight to the top of the tree? Round your answer to the nearest degree.

$75{^o}$

## The Special Angles

The trigonometric ratios for most angles are irrational numbers, but there are a few angles whose trig ratios are “nice” values. You already know one of these values: the sine of $30{^o}\text{.}$ Because the sides of a right triangle are related by the Pythagorean theorem, if we know any one of the trig ratios for an angle, we can find the others. Recall that the side opposite a $30{^o}$ angle is half the length of the hypotenuse, so $\sin 30{^o} = \dfrac{1}{2}\text{.}$

The figure at right shows a 30-60-90 triangle with hypotenuse of length $2$. The opposite side has length 1, and we can calculate the length of the adjacent side. \begin{align*} 1^2 + b^2 = 2^2 \\ b^2 = 2^2 - 1^2 = 3\\ b = \sqrt{3} \end{align*}

Now we know the cosine and tangent of $30{^o}\text{.}$ $\begin{equation*} \cos 30{^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}} = \dfrac{\sqrt{3}}{2} \qquad \tan 30{^o} = \dfrac {\text{opposite}}{\text{adjacent}} = \dfrac{1}{\sqrt{3}} \end{equation*}$

These are exact values for the trig ratios, but we can also find decimal approximations. Use your calculator to verify the following approximate values. \begin{align*} \ \ {\text{exact value}} \ \ {\text{approximation}}\\ \cos 30{^o} = \dfrac{\sqrt{3}}{2} \approx 0.8660\\ \tan 30{^o} = \dfrac{1}{\sqrt{3}} \approx 0.5774 \end{align*}

## Caution 2.39.

It is important for you to understand the difference between exact and approximate values. These decimal approximations, like nearly all the other trig values your calculator gives you, are rounded off. Even if your calculator shows you ten or twelve digits, the values are not exactly correct — although they are quite adequate for most practical calculations!

The angles $30{^o}\text{,}$ $60{^o}\text{,}$ and $45{^o}$ are “special” because we can easily find exact values for their trig ratios and use those exact values to find exact lengths for the sides of triangles with those angles.

Example 2.40.

The sides of an equilateral triangle are $8$ centimeters long. Find the exact length of the triangle’s altitude, $h\text{.}$

The altitude divides the triangle into two 30-60-90 right triangles as shown in the figure. The altitude is adjacent to the $30{^o}$ angle, and the hypotenuse of the right triangle is 8 centimeters. Thus,

\begin{align*} \cos 35 {^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}} \ \ {\text{Fill in the values.}}\\ \dfrac{\sqrt{3}}{2} = \dfrac{h}{8} \ \ {\text{Multiply both sides by 8.}}\\ h = 8\left(\dfrac{\sqrt{3}}{2}\right) = 4\sqrt{3} \end{align*} The altitude is exactly $4\sqrt{3}$ centimeters long.

From this exact answer, we can find approximations to any degree of accuracy we like. You can check that $4\sqrt{3} \approx6.9282\text{,}$ so the altitude is approximately 6.9 centimeters long.

## Checkpoint 2.41.

Use the figure in the previous example to find exact values for the sine, cosine, and tangent of $60{^o}\text{.}$

$\sin 60{^o} = \dfrac{\sqrt{3}}{2}\text{,}$ $\cos 60{^o} = \dfrac{1}{2}\text{,}$ $\tan 60{^o} = \sqrt{3}$

There is one more special angle: $45{^o}\text{.}$ We find the trig ratios for this angle using an isosceles right triangle. Because the base angles of an isosceles triangle are equal, they must both be $45{^o}\text{.}$ The figure shows an isosceles right triangle with equal sides of length 1. You can use the Pythagorean theorem to show that the hypotenuse has length $\sqrt{2}\text{,}$ so the trig ratios for $45{^o}$ are\begin{align*} \sin 45{^o} = \dfrac {\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}} \approx 0.7071\\ \cos 45{^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}} \approx 0.7071\\ \tan 45{^o} = \dfrac {\text{opposite}}{\text{adjacent}} = 1 \end{align*}

## The Trigonometric Ratios for the Special Angles

Here is a summary of the trig ratios for the special angles.

You should memorize the exact values for these trig ratios. A good way to remember them is to know the two special triangles shown below. From these triangles, you can always write down the three trig ratios for the special angles.

You should also be able to recognize their decimal approximations.

We can use the special angles as benchmarks for estimating and mental calculation. For example, we know that $60{^o} \approx 0.8660\text{,}$ so if sin $\theta = 0.95$ for some unknown angle $\theta\text{,}$ we know that $\theta \gt 60{^o}\text{,}$ because as $\theta$ increases from the sine of $\theta$ increases also.

## Example 2.43.

If $\cos \alpha \gt \dfrac{\sqrt{3}}{2}\text{,}$ what can we say about $\alpha\text{?}$

As an angle increases from $0{^o}$ to $90{^o}\text{,}$ its cosine decreases. Now, $\cos 30{^o} = \dfrac{\sqrt{3}}{2}\text{,}$ so if $\cos \alpha \gt \dfrac{\sqrt{3}}{2}\text{,}$ then $\alpha$ must be less than $30{^o}\text{.}$

## Checkpoint 2.44.

If $1 \lt \tan \beta \lt \sqrt{3}\text{,}$ what can we say about $\beta\text{?}$

$45{^o} \lt \beta \lt 60{^o}$

## Section 2.3 Summary

• Solve a triangle
• Inverse sine
• Inverse cosine
• Inverse tangent
• Special angles
• Exact value
• Decimal approximation
• If we know one of the sides of a right triangle and any one of the other four parts, we can use trigonometry to find all the other unknown parts.
• If we know one of the trigonometric ratios of an acute angle, we can find the angle using the inverse trig key on a calculator.
• For the trigonometric ratios of most angles, your calculator gives approximations, not exact values.

## Study Questions

• How many parts of a right triangle (including the right angle) do you need to know in order to solve the triangle?
• Why is it better to use the given values, rather than values you have calculated, when solving a triangle?
• What is the $\sin^{-1}$ (or $\cos^{-1}$ or $\tan^{-1}$) button on the calculator used for?
• Which are the “special” angles, and why are they special?

If we know three parts of a right triangle, including one of the sides, we can use trigonometry to find all the other unknown parts. This is called solving the triangle.

The inverse sine (arcsine) of a value is the angle whose sine is equal to that value.

The inverse cosine (arccosine) of a value is the angle whose sine is equal to that value.

The inverse tangent (arctangent) of a value is the angle whose sine is equal to that value.

Special angles in trigonometry are angles whose sine, cosine, and tangent values can be calculated exactly without the use of a calculator.

An exact value for a trigonometric function is a value that can be represented as a finite combination of integers, radicals, and/or known mathematical constants without the use of a calculator or approximations.

A decimal approximation for a trigonometric function is an estimation of its value expressed as a finite string of decimal digits, usually rounded to a certain number of decimal places, and obtained through the use of a calculator or other computational tool.

## Solve Right Triangle Problems

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## Chapter 4.2: Right Triangle Trigonometry

Using right triangle trigonometry to solve applied problems, using trigonometric functions.

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

## How To: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.

• For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
• Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
• Using the value of the trigonometric function and the known side length, solve for the missing side length.

## Example 5: Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11.

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

We rearrange to solve for $a$.

We can use the sine to find the hypotenuse.

Again, we rearrange to solve for $c$.

A right triangle has one angle of $\frac{\pi }{3}$ and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye.

## How To: Given a tall object, measure its height indirectly.

• Make a sketch of the problem situation to keep track of known and unknown information.
• Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
• At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
• Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
• Solve the equation for the unknown height.

## Example 6: Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of $57^\circ$ between a line of sight to the top of the tree and the ground, as shown in Figure 13. Find the height of the tree.

We know that the angle of elevation is $57^\circ$ and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of $57^\circ$, letting $h$ be the unknown height.

The tree is approximately 46 feet tall.

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of $\frac{5\pi }{12}$ with the ground? Round to the nearest foot.

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## Definition

Historical significance, right angle, pythagorean theorem, special right triangles, trigonometric functions, congruence and similarity, cosine (cos), tangent (tan), special right triangle formulas, area formula, perimeter formula, acute right triangle, obtuse right triangle, isosceles right triangle, engineering, surveying and navigation, architecture and construction, art and design.

The right triangle , a cornerstone of geometry , is a geometric shape that embodies both simplicity and mathematical significance . With its distinctive form defined by one right angle , the right triangle holds a special place in various fields, including mathematics, physics, engineering , and architecture . Its elegant proportions and inherent properties make it a fundamental building block for constructing complex structures and solving intricate problems.

We set off on a tour into the fascinating world of the right triangle in this essay , exploring its unique characteristics , fundamental principles , Pythagorean theorem , and practical applications . By unraveling the secrets of this geometric marvel, we aim to shed light on the profound impact of right triangles in shaping the foundations of theoretical knowledge and real-world structures.

A  right triangle  is a  geometric shape  that is defined by its specific properties. It is a  triangle  that contains one  right angle , which measures  90 degrees . The right angle is formed by the intersection of two sides, known as the  legs , and it is always opposite the longest side of the triangle, called the  hypotenuse .

The other two angles in a right triangle are acute , or less than 90 degrees . The two sharp angles in a right triangle must add up to 90 degrees since the total angle in any triangle is always 180 degrees . Below is the generic diagram for a right triangle.

Figure-1: Generic right triangle.

The Pythagorean theorem is one of the basic concepts related to right triangles. The square of the lengths of the two legs in a right triangle must equal the square of the length of the hypotenuse , according to this theorem. This can be mathematically represented as $c^2 = a^2 + b^2$, “c” being the length of the hypotenuse and “a” and “b” being the lengths of the legs.

The historical background of the  right triangle , a geometric shape, dates back to ancient cultures and was crucial in the growth of mathematics and several scientific fields. The concept of the right triangle can be traced back to  Ancient Mesopotamia  and  Ancient Egypt , where early civilizations recognized its special properties and utilized it in their architectural and surveying practices. The Egyptians, in particular, were skilled in geometry and used right triangles in their construction projects, such as the pyramids.

The study of right triangles flourished in  Ancient Greece , with renowned mathematicians such as  Pythagoras  making significant contributions. Pythagoras’ Theorem states that the hypotenuse’s square is equal to the sum of its two other sides’ squares. according to Pythagoras’ Theorem. became a fundamental principle in geometry and trigonometry. The theorem is named after Pythagoras, but similar ideas were known in other ancient cultures as well.

During the  Islamic Golden Age  (8th to 14th centuries), scholars like  Al-Khwarizmi  and  Ibn al-Haytham  further expanded their knowledge of right triangles and their properties. They contributed to advancements in trigonometry, introducing new concepts and techniques for solving problems involving right triangles.

In  Renaissance Europe , mathematicians and astronomers such as  Leonardo da Vinci ,  Johannes Kepler , and  Galileo Galilei  relied on right triangles and trigonometry to explore the principles of optics, planetary motion, and perspective in art.

The study of right triangles continued to progress through the  Enlightenment  and into the modern era, with the development of trigonometric functions and their applications in various scientific and engineering fields.

Today, the understanding of right triangles and their properties is an essential part of mathematics education . Right triangles find extensive use in fields such as engineering, physics, navigation, computer graphics, and surveying . Trigonometry , which heavily relies on right triangles, provides a comprehensive framework for analyzing and solving problems involving angles, distances , and relationships between sides .

## Fundamental Properties

Certainly! Here are the properties of the  right triangle , a geometric shape, explained in detail:

The most distinguishing property of a right triangle is the presence of a right angle . A right angle measures  90 degrees  and is formed where two sides of the triangle meet to create a perpendicular intersection.

In a right triangle , the hypotenuse is the side that faces the right angle. It is the longest side of the triangle and is denoted by the letter c . The hypotenuse is directly across from the right angle and forms the base for many important relationships in right triangles.

The other two sides of the right triangle , which are adjacent to the right angle, are called the legs . They are denoted by the letters  a  and  b . The lengths of the legs can vary and are crucial for calculating other properties of the triangle.

One of the fundamental properties of a right triangle is the Pythagorean theorem , named after the Greek mathematician Pythagoras.The square of the hypotenuse is equal to the sum of the squares of the two legs, according to this rule. In equation form, it is written as  c² = a² + b² .

Right triangles can have specific ratios between their side lengths, resulting in special right triangles . The two most common special right triangles are the  45-45-90 triangle  and the  30-60-90 triangle . In a 45-45-90 triangle , the two legs are congruent , and the hypotenuse is equal to the length of a leg multiplied by √2 . In a 30-60-90 triangle , the lengths of the sides follow a specific ratio: the length of the shorter leg is half the length of the hypotenuse , and the length of the longer leg is √3 times the length of the shorter leg.

In a 45-45-90 triangle , the two legs are congruent , and the hypotenuse is equal to the length of a leg multiplied by √2 . In a 30-60-90 triangle , the lengths of the sides follow a specific ratio: the length of the shorter leg is half the length of the hypotenuse , and the length of the longer leg is √3 times the length of the shorter leg.

Right triangles can be used to determine whether two triangles are congruent (having identical side lengths and angles) or similar (having proportional side lengths and congruent angles).

Understanding these properties of the right triangle allows for a comprehensive exploration of its geometry , relationships between sides and angles, and practical applications . Right triangles and their properties serve as the basis for trigonometry , navigation , engineering , and countless other fields where accurate calculations involving angles and distances are required.

## Related Formulas

Below are the related formulas of the right triangle , explained in detail:

One of the most important formulas for right triangles is the Pythagorean theorem . The hypotenuse is the side that forms the right angle, and the rule says that the square of its length is equal to the sum of the squares of the lengths of the other two sides. In equation form, it is written as c² = a² + b² , where a and b are the lengths of the two legs, and c is the hypotenuse length.

Trigonometric functions are widely used in right triangles to relate the ratios of the side lengths to the angles within the triangle. The three primary trigonometric functions are:

The sine of an angle in a right triangle is the proportion of the length of the side directly across from the angle to the length of the hypotenuse.It is written as sin(A) = opposite / hypotenuse .

The ratio of the neighboring side’s length to the hypotenuse’s length is known as the cosine of an angle in a right triangle . It is written as cos(A) = adjacent / hypotenuse .

In a right triangle , the lengths of the adjacent and opposing sides are compared to determine the angle’s tangent . It is written as tan(A) = opposite/adjacent .

Special right triangles , such as the 45-45-90 triangle and the 30-60-90 triangle , have specific formulas that relate to their side lengths. In a 45-45-90 triangle , the lengths of the two legs are equal, and the length of the hypotenuse is equal to the length of a leg multiplied by √2 . In a 30-60-90 triangle , the lengths of the sides follow a specific ratio: the length of the shorter leg is half the length of the hypotenuse, and the greater leg is about √3 times longer than the shorter leg.

The formula to calculate the area of a right triangle is A = 0.5 × base × height , where the base and height are the lengths of the two legs. The base and height can be any two sides of the triangle as long as they form a right angle .

The total length of all the sides makes up a right triangle’s perimeter . In a right triangle, the perimeter can be calculated by adding the lengths of the two legs and the hypotenuse: P = a + b + c .

There are three main types of  right triangles  based on the relationships between their sides and angles. Let’s explore each type in detail:

An acute right triangle  is a right triangle where both of the  acute angles  (the angles that are less than 90 degrees) are  less than 45 degrees . In this type of triangle, the two legs are of different lengths, and the hypotenuse is the longest side. The Pythagorean theorem applies to acute right triangles, where c² = a² + b² . The acute right triangle is characterized by its sharp angles and is commonly encountered in mathematical problems and applications. Below is the diagram for an acute right triangle .

Figure-2: Acute right triangle.

An obtuse right triangle is a right triangle where one of the acute angles is greater than 45 degrees . The other acute angle remains less than 45 degrees , and the right angle is always 90 degrees . In an obtuse right triangle, the longest side is the one opposite the obtuse angle . The Pythagorean theorem also applies to obtuse right triangles, following the same formula as the acute right triangle. These triangles possess one larger angle , which gives them a distinct shape and characteristic. Below is the diagram for an obtuse right triangle .

Figure-3: Obtuse right triangle.

An isosceles right triangle is a right triangle where two of the sides, the legs , are of equal length. The remaining side, the hypotenuse , is always longer than the legs. In an isosceles right triangle, both of the acute angles are 45 degrees . The Pythagorean theorem is simplified for isosceles right triangles as c² = 2a² , where “a” denotes the length of the legs and “c” the length of the hypotenuse. Isosceles right triangles exhibit symmetry and are often used in geometric constructions and calculations. Below is the diagram for an isosceles right triangle .

Figure-4: Isosceles right triangle.

Understanding the types of right triangles is important as it provides insights into their specific characteristics and properties . These types allow for a classification of right triangles based on their angle measures and side lengths , enabling more precise analysis and problem-solving in various mathematical and practical contexts .

## Applications

Right triangles are prevalent in engineering disciplines . They are utilized in structural design , architecture , and civil engineering for calculating forces, determining stability, and analyzing structural elements. Right triangles are also essential in fields like electrical engineering and circuit design , where they aid in calculating voltage, resistance, and impedance in complex circuits.

Right triangles find applications in physics for various calculations involving forces , vectors , and motion . In mechanics , right triangles are used to resolve forces into components and calculate their magnitudes and directions. Additionally, right triangles are employed in trigonometric calculations related to projectile motion , circular motion , and waves .

Right triangles play a significant role in surveying and navigation . Triangulation methods , which rely on the principles of right triangles, are used to determine distances , angles , and positions of points on the Earth’s surface. This is crucial in activities such as land surveying , mapmaking , and global positioning systems (GPS) technology .

Right triangles are essential in architectural design and construction . They aid in creating geometrically balanced structures, determining angles for roof pitches, calculating stair dimensions, and ensuring proper alignments and measurements.

Right triangles are utilized in celestial navigation and astronomical calculations . Triangulation methods involving right triangles help determine the positions of celestial objects, measure distances in space, and navigate using celestial coordinates.

Right triangles are employed in art and design compositions to create balanced and aesthetically pleasing arrangements. The principles of right triangles help in achieving visual harmony , proportion , and perspective .

Understanding the applications of the right triangle in various fields provides insights into its practical significance and demonstrates its universal utility in different domains.

Find the length of the hypotenuse in a right triangle with legs measuring 3 units and 4 units.

Given: Length of leg a  =  3 units  Length of leg  b  =  4 units

We can apply the Pythagorean theorem to determine the length of the hypotenuse:  c² = a² + b² .

Substituting the given values:

c² = 3² + 4²

c² = 9 + 16

When we square the two sides, we discover:

c = 5 units

Therefore, the length of the hypotenuse in the right triangle is  5 units .

Determine the length of a leg in a right triangle with a hypotenuse of 10 units and the other leg measuring 6 units.

Given: Length of hypotenuse c  =  10 units  Length of one leg  a  =  6 units

To find the length of the other leg,  b , we can use the Pythagorean theorem:  c² = a² + b² .

10² = 6² + b²

100 = 36 + b²

b² = 100 – 36

b = 8 units

Therefore, the length of the other leg in the right triangle is  8 units .

Calculate the area of a right triangle given in Figure-5.

Given: Base b  =  5 units  Height  h  =  7 units

To find the area  A  of a right triangle, we can use the formula:  A = 0.5 × b × h .

A = 0.5 × 5 × 7

A = 0.5 × 35

A = 17.5 square units

Therefore, the area of the right triangle is  17.5 square units .

Determine the length of a leg in a right triangle with a hypotenuse of 13 units and the other leg measuring 5 units.

Given: Length of hypotenuse c  =  13 units  Length of one leg  a  =  5 units

13² = 5² + b²

169 = 25 + b²

b² = 169 – 25

b = 12 units

Therefore, the length of the other leg in the right triangle is  12 units .

Find the measure of an acute angle in a right triangle if the lengths of the legs are 3 units and 4 units.

To find the measure of an acute angle, we can use trigonometric functions. Let’s calculate the sine of the angle.

Using the formula:  sin(A) = opposite / hypotenuse : sin(A) = 3 / 5

Taking the inverse sine (sin⁻¹) of both sides, we find:

A = sin⁻¹(3 / 5)

A ≈ 36.87 degrees

Therefore, the measure of the acute angle in the right triangle is approximately  36.87 degrees .

Calculate the perimeter of a right triangle with legs measuring 6 units and 8 units.

Given: Length of leg a  =  6 units  Length of leg  b  =  8 units

To find the perimeter  P  of a right triangle, we can add the lengths of all sides:  P = a + b + c .

Since the perimeter is not given directly, the hypotenuse’s length must be determined, using the Pythagorean theorem:  c² = a² + b² .

c² = 6² + 8²

c² = 36 + 64

c = 10 units

Now we can calculate the perimeter:

P = 6 + 8 + 10

P = 24 units

Therefore, the perimeter of the right triangle is  24 units .

Determine the length of the hypotenuse in a right triangle, with one leg measuring 9 units and the other leg measuring 12 units.

Given: Length of leg a  =  9 units  Length of leg  b  =  12 units

We can apply the Pythagorean theorem to determine the length of the hypotenuse c::  c² = a² + b² .

c² = 9² + 12²

c² = 81 + 144

c = 15 units

Therefore, the length of the hypotenuse in the right triangle is  15 units .

Find the area of a right triangle with a base of 12 units and a height of 5 units.

Given: Base b  =  12 units  Height  h  =  5 units

A = 0.5 × 12 × 5

A = 0.5 × 60

A = 30 square units

Therefore, the area of the right triangle is  30 square units .

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## Right-Triangle Word Problems

What is a right-triangle word problem.

A right-triangle word problem is one in which you are given a situation (like measuring something's height) that can be modelled by a right triangle. You will draw the triangle, label it, and then solve it; finally, you interpret this solution within the context of the original exercise.

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## MathHelp.com

Right Triangle Word Problems

Once you've learned about trigonometric ratios (and their inverses), you can solve triangles. Naturally, many of these triangles will be presented in the context of word problems. A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it. Once you've got a helpful diagram, the math is usually pretty straightforward.

• A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the ground, how far up the wall does the ladder reach? How far from the wall is the base of the ladder? Round your answers to two decimal places, as needed.

First, I'll draw a picture. It doesn't have to be good or to scale; it just needs to be clear enough that I can keep track of what I'm doing. My picture is:

To figure out how high up the wall the top of the ladder is, I need to find the height h of my triangle.

Since they've given me an angle measure and "opposite" and the hypotenuse for this angle, I'll use the sine ratio for finding the height:

sin(60°) = h/6

6 sin(60°) = h = 3sqrt[3]

Plugging this into my calculator, I get an approximate value of 5.196152423 , which I'll need to remember to round when I give my final answer.

For the base, I'll use the cosine ratio:

cos(60°) = b/6

6×cos(60°) = b = 3

Nice! The answer is a whole number; no radicals involved. I won't need to round this value when I give my final answer. Checking the original exercise, I see that the units are "meters", so I'll include this unit on my numerical answers:

Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting. For instance, if they hadn't told me to round my numbers in the exercise above, my value for the height would have been the value with the radical.

• A five-meter-long ladder leans against a wall, with the top of the ladder being four meters above the ground. What is the approximate angle that the ladder makes with the ground? Round to the nearest whole degree.

As usual, I'll start with a picture, using "alpha" to stand for the base angle:

They've given me the "opposite" and the hypotenuse, and asked me for the angle value. For this, I'll need to use inverse trig ratios.

sin(α) = 4/5

m(α) = sin −1 (4/5) = 53.13010235...

(Remember that m(α) means "the measure of the angle α".)

So I've got a value for the measure of the base angle. Checking the original exercise, I see that I am supposed to round to the nearest whole degree, so my answer is:

base angle: 53°

• You use a transit to measure the angle of the sun in the sky; the sun fills 34' of arc. Assuming the sun is 92,919,800 miles away, find the diameter of the sun. Round your answer to the nearest whole mile.

First, I'll draw a picture, labelling the angle on the Earth as being 34 minutes, where minutes are one-sixtieth of a degree. My drawing is *not* to scale!:

Hmm... This "ice-cream cone" picture doesn't give me much to work with, and there's no right triangle.

The two lines along the side of my triangle measure the lines of sight from Earth to the sides of the Sun. What if I add another line, being the direct line from Earth to the center of the Sun?

Now that I've got this added line, I have a right triangle — two right triangles, actually — but I only need one. I'll use the triangle on the right.

(The angle measure , "thirty-four arc minutes", is equal to 34/60 degrees. Dividing this in half is how I got 17/60 of a degree for the smaller angle.)

I need to find the width of the Sun. That width will be twice the base of one of the right triangles. With respect to my angle, they've given me the "adjacent" and have asked for the "opposite", so I'll use the tangent ratio:

tan(17/60°) = b/92919800

92919800×tan(17/60°) = b = 459501.4065...

This is just half the width; carrying the calculations in my calculator (to minimize round-off error), I get a value of 919002.8129 . This is higher than the actual diameter, which is closer to 864,900 miles, but this value will suffice for the purposes of this exercise.

• A private plane flies 1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues another 1.5 hours at the same speed, but on a bearing of 130°. At the end of this time, how far is the plane from its starting point? What is its bearing from that starting point? Round your answers to whole numbers.

The bearings tell me the angles from "due north", in a clockwise direction. Since 130 − 40 = 90 , these two bearings create a right angle where the plane turns. From the times and rates, I can find the distances travelled in each part of the trip:

1.3 × 110 = 143 1.5 × 110 = 165

Now that I have the lengths of the two legs, I can set up a triangle:

(The angle θ is the bearing, from the starting point, of the plane's location at the ending point of the exercise.)

I can find the distance between the starting and ending points by using the Pythagorean Theorem :

143 2 + 165 2 = c 2 20449 + 27225 = c 2 47674 = c 2 c = 218.3437657...

The 165 is opposite the unknown angle, and the 143 is adjacent, so I'll use the inverse of the tangent ratio to find the angle's measure:

165/143 = tan(θ)

tan −1 (165/143) = θ = 49.08561678...

But this angle measure is not the "bearing" for which they've asked me, because the bearing is the angle with respect to due north. To get the measure they're wanting, I need to add back in the original forty-degree angle:

distance: 218 miles

bearing: 89°

Related: Another major class of right-triangle word problems you will likely encounter is angles of elevation and declination .

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## Similar Right Triangles Fully Explained w/ 9 Examples!

// Last Updated: January 21, 2020 - Watch Video //

In today’s geometry lesson, you’re going to learn all about similar right triangles.

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

More specifically, you’re going to see how to use the geometric mean to create proportions, which in turn help us solve for missing side lengths.

Let’s get started!

How are right triangles and the geometric mean related?

The two legs meet at a 90° angle, and the hypotenuse is the side opposite the right angle and is the longest side.

Right Triangle Diagram

The geometric mean of two positive numbers a and b is:

Geometric Mean of Two Numbers

And the geometric mean helps us find the altitude of a right triangle! In fact, the geometric mean, or mean proportionals, appears in two critical theorems on right triangles.

## Geometric Mean Theorems

In a right triangle, if the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments, then the length of the altitude is the geometric mean of the lengths of the two segments.

Altitude Rule

Additionally, the length of each leg is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg, as ck-12 accurately states.

But what do these theorems really mean?

They help us to create proportions for finding missing side lengths!

Let’s look at an example!

## How To Solve Similar Right Triangles

In the figure below, we are being asked to find the altitude, using the geometric mean and the given lengths of two segments:

Using Similar Right Triangles

In the video below, you’ll learn how to deal with harder problems, including how to solve for the three different types of problems:

• Missing Altitude
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## 10.9: Right Triangle Trigonometry

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## Learning Objectives

After completing this section, you should be able to:

• Apply the Pythagorean Theorem to find the missing sides of a right triangle.
• Apply the 30 ∘ - 60 ∘ - 90 ∘ 30 ∘ - 60 ∘ - 90 ∘ and 45 ∘ - 45 ∘ - 90 ∘ 45 ∘ - 45 ∘ - 90 ∘ right triangle relationships to find the missing sides of a triangle.
• Apply trigonometric ratios to find missing parts of a right triangle.
• Solve application problems involving trigonometric ratios.

This is another excerpt from Raphael’s The School of Athens. The man writing in the book represents Pythagoras, the namesake of one of the most widely used formulas in geometry, engineering, architecture, and many other fields, the Pythagorean Theorem. However, there is evidence that the theorem was known as early as 1900–1100 BC by the Babylonians. The Pythagorean Theorem is a formula used for finding the lengths of the sides of right triangles.

Born in Greece, Pythagoras lived from 569–500 BC. He initiated a cult-like group called the Pythagoreans, which was a secret society composed of mathematicians, philosophers, and musicians. Pythagoras believed that everything in the world could be explained through numbers. Besides the Pythagorean Theorem, Pythagoras and his followers are credited with the discovery of irrational numbers, the musical scale, the relationship between music and mathematics, and many other concepts that left an immeasurable influence on future mathematicians and scientists.

The focus of this section is on right triangles. We will look at how the Pythagorean Theorem is used to find the unknown sides of a right triangle, and we will also study the special triangles, those with set ratios between the lengths of sides. By ratios we mean the relationship of one side to another side. When you think about ratios, you should think about fractions. A fraction is a ratio, the ratio of the numerator to the denominator. Finally, we will preview trigonometry. We will learn about the basic trigonometric functions, sine, cosine and tangent, and how they are used to find not only unknown sides but unknown angles, as well, with little information.

## Pythagorean Theorem

The Pythagorean Theorem is used to find unknown sides of right triangles. The theorem states that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse (the longest side of the right triangle).

The Pythagorean Theorem states

a 2 + b 2 = c 2 a 2 + b 2 = c 2

where a Figure 10.179.

For example, given that side a = 6 , a = 6 , and side b = 8 , b = 8 , we can find the measure of side c c using the Pythagorean Theorem. Thus,

## Example 10.64

Using the pythagorean theorem.

Find the length of the missing side of the triangle (Figure 10.180).

Using the Pythagorean Theorem, we have

( 6 ) 2 + b 2 = ( 14 ) 2 36 + b 2 = 196 b 2 = 196 − 36 b 2 = 160 b = ± 160 = 4 10 = 12.65 ( 6 ) 2 + b 2 = ( 14 ) 2 36 + b 2 = 196 b 2 = 196 − 36 b 2 = 160 b = ± 160 = 4 10 = 12.65

When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root.

The applications of the Pythagorean Theorem are countless, but one especially useful application is that of distance. In fact, the distance formula stems directly from the theorem. It works like this:

In Figure 10.182, the problem is to find the distance between the points ( − 3 , − 1 ) ( − 3 , − 1 ) and ( 3 , 2 ) . ( 3 , 2 ) . We call the length from point ( − 3 , − 1 ) ( − 3 , − 1 ) to point ( 3 , − 1 ) ( 3 , − 1 ) side a a , and the length from point ( 3 , − 1 ) ( 3 , − 1 ) to point ( 3 , 2 ) ( 3 , 2 ) side b b . To find side c c , we use the distance formula and we will explain it relative to the Pythagorean Theorem. The distance formula is d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 , d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 , such that ( x 2 − x 1 ) ( x 2 − x 1 ) is a substitute for a a in the Pythagorean Theorem and is equal to 3 − ( − 3 ) = 6 ; 3 − ( − 3 ) = 6 ; and ( y 2 − y 1 ) ( y 2 − y 1 ) is a substitute for b b in the Pythagorean Theorem and is equal to 2 − ( − 1 ) = 3. 2 − ( − 1 ) = 3. When we plug in these numbers to the distance formula, we have

d = ( 3 − ( − 3 ) ) 2 + ( 2 − ( − 1 ) ) 2 = ( 6 ) 2 + ( 3 ) 2 = 36 + 9 = 45 = 3 5 = 6.7 d = ( 3 − ( − 3 ) ) 2 + ( 2 − ( − 1 ) ) 2 = ( 6 ) 2 + ( 3 ) 2 = 36 + 9 = 45 = 3 5 = 6.7

Thus, d = c d = c , the hypotenuse, in the Pythagorean Theorem.

## Example 10.65

Calculating distance using the distance formula.

You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive (Figure 10.183). You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft.

You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then 9 ( 200 ) + 7 ( 200 ) = 1,800 ft + 1,400 ft = 3,200 ft 9 ( 200 ) + 7 ( 200 ) = 1,800 ft + 1,400 ft = 3,200 ft .

As the crow flies, use the distance formula. We have

d = ( 1,800 − 0 ) 2 + ( 1,400 − 0 ) 2 = 3,240,000 + 1,960,000 = 5,200,000 = 2280.4 ft d = ( 1,800 − 0 ) 2 + ( 1,400 − 0 ) 2 = 3,240,000 + 1,960,000 = 5,200,000 = 2280.4 ft

Example 10.66, calculating distance with the pythagorean theorem.

The city has specific building codes for wheelchair ramps. Every vertical rise of 1 in requires that the horizontal length be 12 inches. You are constructing a ramp at your business. The plan is to make the ramp 130 inches in horizontal length and the slanted distance will measure approximately 132.4 inches (Figure 10.184). What should the vertical height be?

The Pythagorean Theorem states that the horizontal length of the base of the ramp, side a , is 130 in. The length of c , or the length of the hypotenuse, is 132.4 in. The length of the height of the triangle is side b .

Then, by the Pythagorean Theorem, we have:

a 2 + b 2 = c 2 ( 130 ) 2 + b 2 = ( 132.4 ) 2 16,900 + b 2 = 17,529.76 b 2 = 17,529.76 − 16,900 b 2 = 629.8 b = 629.8 = 25 a 2 + b 2 = c 2 ( 130 ) 2 + b 2 = ( 132.4 ) 2 16,900 + b 2 = 17,529.76 b 2 = 17,529.76 − 16,900 b 2 = 629.8 b = 629.8 = 25

If you construct the ramp with a 25 in vertical rise, will it fulfill the building code? If not, what will have to change?

The building code states 12 in of horizontal length for each 1 in of vertical rise. The vertical rise is 25 in, which means that the horizontal length has to be 12 ( 25 ) = 300 in . 12 ( 25 ) = 300 in . So, no, this will not pass the code. If you must keep the vertical rise at 25 in, what will the other dimensions have to be? Since we need a minimum of 300 in for the horizontal length:

( 300 ) 2 + ( 25 ) 2 = c 2 90,625 = c 2 90,625 = c = 301 in ( 300 ) 2 + ( 25 ) 2 = c 2 90,625 = c 2 90,625 = c = 301 in

The new ramp will look like Figure 10.185.

## 30 ∘ - 60 ∘ - 90 ∘ 30 ∘ - 60 ∘ - 90 ∘ Triangles

In geometry, as in all fields of mathematics, there are always special rules for special circumstances. An example is the perfect square rule in algebra. When expanding an expression like ( 2 x + 5 y ) 2 , ( 2 x + 5 y ) 2 , we do not have to expand it the long way:

( 2 x + 5 y ) 2 = ( 2 x + 5 y ) ( 2 x + 5 y ) = ( 2 x ) 2 + 10 x y + 10 x y + ( 5 y ) 2 = 4 x 2 + 20 x y + 25 y 2 ( 2 x + 5 y ) 2 = ( 2 x + 5 y ) ( 2 x + 5 y ) = ( 2 x ) 2 + 10 x y + 10 x y + ( 5 y ) 2 = 4 x 2 + 20 x y + 25 y 2

If we know the perfect square formula, given as

( a + b ) 2 = a 2 + 2 a b + b 2 , ( a + b ) 2 = a 2 + 2 a b + b 2 ,

we can skip the middle step and just start writing down the answer. This may seem trivial with problems like ( a + b ) 2 . Figure 10.187.

We see that the shortest side is opposite the smallest angle, and the longest side, the hypotenuse, will always be opposite the right angle. There is a set ratio of one side to another side for the 30 ∘ - 60 ∘ - 90 ∘ 30 ∘ - 60 ∘ - 90 ∘ triangle given as 1 : 3 : 2 , 1 : 3 : 2 , or x : x 3 : 2 x . x : x 3 : 2 x . Thus, you only need to know the length of one side to find the other two sides in a 30 ∘ - 60 ∘ - 90 ∘ 30 ∘ - 60 ∘ - 90 ∘ triangle.

## Example 10.67

Finding missing lengths in a 3 0 ∘ - 6 0 ∘ - 9 0 ∘ 3 0 ∘ - 6 0 ∘ - 9 0 ∘ triangle.

Find the measures of the missing lengths of the triangle (Figure 10.188).

We can see that this is a 30 ∘ - 60 ∘ - 90 ∘ 30 ∘ - 60 ∘ - 90 ∘ triangle because we have a right angle and a 30 ∘ 30 ∘ angle. The remaining angle, therefore, must equal 60 ∘ . 60 ∘ . Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side a a is the shortest side, as it is opposite the smallest angle 30 ∘ , 30 ∘ , and we can substitute a = x . a = x . The ratios are x : x 3 : 2 x . x : x 3 : 2 x . We have the hypotenuse equaling 10, which corresponds to side c c , and side c c is equal to 2 x x . Now, we must solve for x x :

2 x = 10 x = 5 2 x = 10 x = 5

Side b b is equal to x 3 x 3 or 5 3 . 5 3 . The lengths are 5 , 5 3 , 10. 5 , 5 3 , 10.

## Example 10.68

Applying 3 0 ∘ - 6 0 ∘ - 9 0 ∘ 3 0 ∘ - 6 0 ∘ - 9 0 ∘ triangle to the real world.

A city worker leans a 40-foot ladder up against a building at a 30 ∘ Figure 10.190). How far up the building does the ladder reach?

We have a 30 ∘ - 60 ∘ - 90 ∘ 30 ∘ - 60 ∘ - 90 ∘ triangle, and the hypotenuse is 40 ft. This length is equal to 2 x x , where x x is the shortest side. If 2 x = 40 2 x = 40 , then x = 20 x = 20 . The ladder is leaning on the wall 20 ft up from the ground.

## 4 5 ∘ - 4 5 ∘ - 9 0 ∘ 4 5 ∘ - 4 5 ∘ - 9 0 ∘ Triangles

The 45 ∘ - 45 ∘ - 90 ∘ Figure 10.192.

## Example 10.69

Finding missing lengths of a 4 5 ∘ - 4 5 ∘ - 9 0 ∘ 4 5 ∘ - 4 5 ∘ - 9 0 ∘ triangle.

Find the measures of the unknown sides in the triangle (Figure 10.193).

Because we have a 45 ∘ - 45 ∘ - 90 ∘ 45 ∘ - 45 ∘ - 90 ∘ triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and 2 . 2 . One leg measures 3, so the other leg, a a , measures 3. Remember the ratio of x : x : x 2 . x : x : x 2 . Then, the hypotenuse, c c , equals 3 2 . 3 2 .

## Trigonometry Functions

Trigonometry developed around 200 BC from a need to determine distances and to calculate the measures of angles in the fields of astronomy and surveying. Trigonometry is about the relationships (or ratios) of angle measurements to side lengths in primarily right triangles. However, trigonometry is useful in calculating missing side lengths and angles in other triangles and many applications.

NOTE: You will need either a scientific calculator or a graphing calculator for this section. It must have the capability to calculate trigonometric functions and express angles in degrees.

Trigonometry is based on three functions. We title these functions using the following abbreviations:

• sin = sine sin = sine
• cos = cosine cos = cosine
• tan = tangent tan = tangent

Letting r = x 2 + y 2 , Table 10.1. The functions are given in terms of x x , y y , and r r , and in terms of sides relative to the angle, like opposite, adjacent, and the hypotenuse.

We will be applying the sine function, cosine function, and tangent function to find side lengths and angle measurements for triangles we cannot solve using any of the techniques we have studied to this point. In Figure 10.195, we have an illustration mainly to identify r r and the sides labeled x x and y y .

An angle θ Figure 10.196, we will solve for the missing sides.

Let’s use the trigonometric functions to find the sides x x and y y . As long as your calculator mode is set to degrees, you do not have to enter the degree symbol. First, let’s solve for y y .

We have sin θ = y r , sin θ = y r , and θ = 60 ∘ . θ = 60 ∘ . Then,

sin 60 ∘ = y 2 2 sin 60 ∘ = y 1.732 = y 3 = y sin 60 ∘ = y 2 2 sin 60 ∘ = y 1.732 = y 3 = y

Next, let’s find x x . This is the cosine function. We have cos θ = x r . cos θ = x r . Then,

cos 60 ∘ = x 2 2 cos 60 ∘ = x = 1 cos 60 ∘ = x 2 2 cos 60 ∘ = x = 1

Now we have all sides, 1 , 3 , 2. Table 10.2 is a list of common angles, which you should find helpful.

## Example 10.70

Using trigonometric functions.

Find the lengths of the missing sides for the triangle (Figure 10.197).

We have a 55 ∘ 55 ∘ angle, and the length of the triangle on the x x -axis is 6 units.

Step 1: To find the length of r r , we can use the cosine function, as cos θ = x r . cos θ = x r . We manipulate this equation a bit to solve for r r :

cos ( 55 ∘ ) = 6 r r cos ( 55 ∘ ) = 6 r = 6 cos ( 55 ∘ ) r = 6 0.5736 = 10.46 cos ( 55 ∘ ) = 6 r r cos ( 55 ∘ ) = 6 r = 6 cos ( 55 ∘ ) r = 6 0.5736 = 10.46

Step 2: We can use the Pythagorean Theorem to find the length of y y . Prove that your answers are correct by using other trigonometric ratios:

6 2 + y 2 = 10.46 2 y 2 = 109.4 − 36 y = 8.57 6 2 + y 2 = 10.46 2 y 2 = 109.4 − 36 y = 8.57

Step 3: Now that we have y y , we can use the sine function to prove that r r is correct. We have sin θ = y r . sin θ = y r .

sin ( 55 ∘ ) = 8.57 r r sin ( 55 ∘ ) = 8.57 r = 8.57 sin ( 55 ∘ ) = 8.57 0.819 = 10.46 sin ( 55 ∘ ) = 8.57 r r sin ( 55 ∘ ) = 8.57 r = 8.57 sin ( 55 ∘ ) = 8.57 0.819 = 10.46

To find angle measurements when we have two side measurements, we use the inverse trigonometric functions symbolized as sin − 1 , sin − 1 , cos − 1 , cos − 1 , or tan − 1 . tan − 1 . The –1 looks like an exponent, but it means inverse. For example, in the previous example, we had x = 6 x = 6 and r = 10.46. r = 10.46. To find what angle has these values, enter the values for the inverse cosine function cos − 1 ( x r ) cos − 1 ( x r ) in your calculator:

cos − 1 ( 6 10.46 ) = 55 ∘ . cos − 1 ( 6 10.46 ) = 55 ∘ .

You can also use the inverse sine function and enter the values of sin − 1 ( y r ) sin − 1 ( y r ) in your calculator given y = 8.57 y = 8.57 and r = 10.46. r = 10.46. We have

sin − 1 ( 8.57 10.46 ) = 55 ∘ . sin − 1 ( 8.57 10.46 ) = 55 ∘ .

Finally, we can also use the inverse tangent function. Recall tan θ = y x . tan θ = y x . We have

tan − 1 ( 8.57 6 ) = 55 ∘ . tan − 1 ( 8.57 6 ) = 55 ∘ .

## Example 10.71

Solving for lengths in a right triangle.

Solve for the lengths of a right triangle in which θ = 30 ∘ Figure 10.199).

Step 1: To find side a a , we use the sine function:

sin 30 ∘ = a 6 6 sin 30 ∘ = a = 3 sin 30 ∘ = a 6 6 sin 30 ∘ = a = 3

Step 2: To find b b , we use the cosine function:

cos 30 ∘ = b 6 6 cos 30 ∘ = b = 5.196 cos 30 ∘ = b 6 6 cos 30 ∘ = b = 5.196

Step 3: Since this is a 30 ∘ - 60 ∘ - 90 ∘ 30 ∘ - 60 ∘ - 90 ∘ triangle and side b b should equal x 3 , x 3 , if we input 3 for x x , we have b = 3 3 . b = 3 3 . Put this in your calculator and you will get 3 3 = 5.196. 3 3 = 5.196.

## Example 10.72

Finding altitude.

A small plane takes off from an airport at an angle of 31.3 ∘ Figure 10.201). If the plane continues that angle of ascent, find its altitude when it is above the peak, and how far it will be above the peak.

To solve this problem, we use the tangent function:

tan 31.3 ∘ = x 3,520 3,520 tan 31.3 ∘ = 2,140 tan 31.3 ∘ = x 3,520 3,520 tan 31.3 ∘ = 2,140

The plane’s altitude when passing over the peak is 2,140 ft, and it is 1,040 ft above the peak.

Example 10.73, finding unknown sides and angles.

Suppose you have two known sides, but do not know the measure of any angles except for the right angle (Figure 10.202). Find the measure of the unknown angles and the third side.

Step 1: We can find the third side using the Pythagorean Theorem:

6 2 + 4 2 = c 2 52 = c 2 2 13 = c 6 2 + 4 2 = c 2 52 = c 2 2 13 = c

Now, we have all three sides.

Step 2: To find θ , θ , we will first find sin θ . sin θ .

sin θ = o p p h y p = 4 2 13 = 2 13 . sin θ = o p p h y p = 4 2 13 = 2 13 .

The angle θ θ is the angle whose sine is 2 13 . 2 13 .

Step 3: To find θ θ , we use the inverse sine function:

θ = sin − 1 ( 2 13 ) = 33.7 ∘ θ = sin − 1 ( 2 13 ) = 33.7 ∘

Step 4: To find the last angle, we just subtract: 180 ∘ − 90 ∘ − 33.7 ∘ = 56.3 ∘ 180 ∘ − 90 ∘ − 33.7 ∘ = 56.3 ∘ .

## Angle of Elevation and Angle of Depression

Other problems that involve trigonometric functions include calculating the angle of elevation and the angle of depression . These are very common applications in everyday life. The angle of elevation is the angle formed by a horizontal line and the line of sight from an observer to some object at a higher level. The angle of depression is the angle formed by a horizontal line and the line of sight from an observer to an object at a lower level.

## Example 10.74

Finding the angle of elevation.

A guy wire of length 110 meters runs from the top of an antenna to the ground (Figure 10.204). If the angle of elevation of an observer to the top of the antenna is 43 ∘ , 43 ∘ , how high is the antenna?

We are looking for the height of the tower. This corresponds to the y y -value, so we will use the sine function:

sin 43 ∘ = y 110 110 sin 43 ∘ = y 75 = y sin 43 ∘ = y 110 110 sin 43 ∘ = y 75 = y

The tower is 75 m high.

## Example 10.75

Finding angle of elevation.

You are sitting on the grass flying a kite on a 50-foot string (Figure 10.206). The angle of elevation is 60 ∘ . 60 ∘ . How high above the ground is the kite?

We can solve this using the sine function, sin θ = o p p h y p . sin θ = o p p h y p .

sin 60 ∘ = x 50 50 sin 60 ∘ = x = 43.3 ft sin 60 ∘ = x 50 50 sin 60 ∘ = x = 43.3 ft

## People in Mathematics

Pythagoras and the pythagoreans.

The Pythagorean Theorem is so widely used that most people assume that Pythagoras (570–490 BC) discovered it. The philosopher and mathematician uncovered evidence of the right triangle concepts in the teachings of the Babylonians dating around 1900 BC. However, it was Pythagoras who found countless applications of the theorem leading to advances in geometry, architecture, astronomy, and engineering.

Among his accolades, Pythagoras founded a school for the study of mathematics and music. Students were called the Pythagoreans, and the school’s teachings could be classified as a religious indoctrination just as much as an academic experience. Pythagoras believed that spirituality and science coexist, that the intellectual mind is superior to the senses, and that intuition should be honored over observation.

Pythagoras was convinced that the universe could be defined by numbers, and that the natural world was based on mathematics. His primary belief was All is Number. He even attributed certain qualities to certain numbers, such as the number 8 represented justice and the number 7 represented wisdom. There was a quasi-mythology that surrounded Pythagoras. His followers thought that he was more of a spiritual being, a sort of mystic that was all-knowing and could travel through time and space. Some believed that Pythagoras had mystical powers, although these beliefs were never substantiated.

Pythagoras and his followers contributed more ideas to the field of mathematics, music, and astronomy besides the Pythagorean Theorem. The Pythagoreans are credited with the discovery of irrational numbers and of proving that the morning star was the planet Venus and not a star at all. They are also credited with the discovery of the musical scale and that different strings made different sounds based on their length. Some other concepts attributed to the Pythagoreans include the properties relating to triangles other than the right triangle, one of which is that the sum of the interior angles of a triangle equals 180 ∘ . 180 ∘ . These geometric principles, proposed by the Pythagoreans, were proven 200 years later by Euclid.

## A Visualization of the Pythagorean Theorem

In Figure 10.208, which is one of the more popular visualizations of the Pythagorean Theorem, we see that square a a is attached to side a a ; square b b is attached to side b b ; and the largest square, square c c , is attached to side c c . Side a a measures 3 cm in length, side b b measures 4 cm in length, and side c c measures 5 cm in length. By definition, the area of square a a measures 9 square units, the area of square b b measures 16 square units, and the area of square c c measures 25 square units. Substitute the values given for the areas of the three squares into the Pythagorean Theorem and we have

Thus, the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse, as stated in the Pythagorean Theorem.

## Section 10.8 Exercises

#### IMAGES

1. How To Solve A Right Triangle For Abc : Right Triangles

2. Solve Right Triangles: Part 1 The Basics

3. Problems involving right triangle trigonometry

4. Trigonometry

5. How To Solve Hard SAT Math Problems

6. How to Solve a Right Triangle in a Word Problem

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1. 4.5 Objective 1

2. GEOMETRY TRIANGLES PROBLEM #reels #maths #trending

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1. Right triangles & trigonometry

Trigonometry 4 units · 36 skills. Unit 1 Right triangles & trigonometry. Unit 2 Trigonometric functions. Unit 3 Non-right triangles & trigonometry. Unit 4 Trigonometric equations and identities. Course challenge. Test your knowledge of the skills in this course. Start Course challenge. Math.

2. Right Triangles, Hypotenuse, Pythagorean Theorem Examples and Practice

A Right Triangle's Hypotenuse. The hypotenuse is the largest side in a right triangle and is always opposite the right angle. (Only right triangles have a hypotenuse ). The other two sides of the triangle, AC and CB are referred to as the 'legs'. In the triangle above, the hypotenuse is the side AB which is opposite the right angle, ∠C ∠ C .

3. 1.4: Solving Right Triangles

Video: Example: Determine What Trig Function Relates Specific Sides of a Right Triangle Practice: Angles of Elevation and Depression This page titled 1.4: Solving Right Triangles is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation .

4. 5.2: Solution of Right Triangles

Example $$\PageIndex{2}$$ Find $$x$$ to the nearest tenth: Solution. We wish to find the leg opposite $$20^{\circ}$$ and we know the hypotenuse. We use the sine because it is the only one of the three trigonometric functions which involves both the opposite leg and the hypotenuse.

5. Solving Right Triangles: Problems

Problem : Solve the following right triangle, in which C = 90 o: A = 40, B = 50. This triangle cannot be solved. Three angles is not sufficient information to determine a unique triangle. Problem : Solve the following right triangle, in which C = 90 o: b = 6, B = 72 o. A = 90 o - B = 18 o.

6. Right Triangle

A right triangle is a type of triangle that has one right angle ( 90^ {\circ} 90∘ angle). The two sides of the triangle that meet at the right angle are perpendicular. In this case, side AB AB is perpendicular to side BC BC so angle B B is 90^ {\circ}. 90∘. A right angle is 90^ {\circ} 90∘ and it is symbolized using a small square inside ...

7. Solving Right Triangles

Example 1: Solve the right triangle shown in Figure (b) if ∠ B = 22°. Because the three angles of a triangle must add up to 180°, ∠ A = 90 ∠ B thus ∠ A = 68°. The following is an alternate way to solve for sides a and c: This alternate solution may be easier because no division is involved. Example 2: Solve the right triangle shown ...

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Recall that the side opposite a 30o 30 o angle is half the length of the hypotenuse, so sin30o = 1 2. sin. ⁡. 30 o = 1 2. The figure at right shows a 30-60-90 triangle with hypotenuse of length 2 2. The opposite side has length 1, and we can calculate the length of the adjacent side. 12 + b2 = 22 b2 = 22 −12 = 3 b = √3 1 2 + b 2 = 2 2 b 2 ...

9. Solve Right Triangle Problems

Right triangles problems are solved and detailed explanations are included. Example - Problem 1: Find sin(x) and cos(x) in the right triangle shown below. Solution to Problem 1: First use the Pythagorean theorem to find the hypotenuse h of the right triangle. h = √(6 2 + 8 2) = √(36 + 64) = 10 ; In a right triangle, using trigonometric ratio for sin(x) we write. sin(x) = 8 / 10 = 0.8

10. Solving Right Triangle Problems

Example 18: Using the Pythagorean Theorem to Model an Equation. Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem. One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced.

11. 1.3: Applications and Solving Right Triangles

In applied problems it is not always obvious which right triangle to use, which is why these sorts of problems can be difficult. Often no right triangle will be immediately evident, so you will have to create one. ... Example 1.19. Solve the right triangle in Figure 1.3.3 using the given information: Figure 1.3.3 (a) $$c = 10,\, A = 22^$$

12. PDF 1. SOLVING RIGHT TRIANGLES Example Solve for x y E

2. APPLICATION PROBLEMS 1. SOLVING RIGHT TRIANGLES Example Solve for x, y, and E. 40q y 6 x To solve for : Since the three angles of any triangle sum to 0q, we get the following equation to solve. E 40 q 90 q 0 q E 40 q 90 q E 50 q Recall: Two angles that sum to 90q are called complimentary angles. The two acute angles in a right triangle are ...

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How To: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides. For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator. Write an equation setting the function value ...

14. The Right Triangle

Figure-4: Isosceles right triangle. Understanding the types of right triangles is important as it provides insights into their specific characteristics and properties.These types allow for a classification of right triangles based on their angle measures and side lengths, enabling more precise analysis and problem-solving in various mathematical and practical contexts.

15. Solving right-triangle word problems: Learn here!

bearing: 89°. Related: Another major class of right-triangle word problems you will likely encounter is angles of elevation and declination. To solve a right-triangle word problem, first read the entire exercise. Draw a right triangle; it need not be 'to scale'. Then start labelling.

16. Right triangle trigonometry word problems

A dashed line connects from the point to the surface above the location to the left of the point. This forms the hypotenuse of a right triangle that is eighty meters in distance. The location to the left of the point up to the surface is the height of the right triangle. The angle opposite the height is unknown.

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Solving a right triangle can be accomplished by using the definitions of the trigonometric functions and the Pythagorean Theorem. This process is called solving a right triangle. Being able to solve a right triangle is useful in solving a variety of real-world problems such as the construction of a wheelchair ramp.

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19. 1.2: Special Right Triangles

30-60-90 Right Triangles. Hypotenuse equals twice the smallest leg, while the larger leg is 3-√ 3 times the smallest. One of the two special right triangles is called a 30-60-90 triangle, after its three angles. 30-60-90 Theorem: If a triangle has angle measures 30∘ 30 ∘, 60∘ 60 ∘ and 90∘ 90 ∘, then the sides are in the ratio x ...

20. Similar Right Triangles (Fully Explained w/ 9 Examples!)

Similar Right Triangles. Fully Explained w/ 9 Examples! In today's geometry lesson, you're going to learn all about similar right triangles. More specifically, you're going to see how to use the geometric mean to create proportions, which in turn help us solve for missing side lengths. Let's get started!

21. 10.9: Right Triangle Trigonometry

Learn how to use the basic trigonometric functions to find the missing angles and sides of right triangles, and how to apply them to real-world problems. This section covers the definitions, properties, and examples of sine, cosine, and tangent functions, as well as the Pythagorean theorem and inverse trigonometric functions. - Mathematics LibreTexts