step in problem solving involving trigonometric ratios

Problems on Trigonometric Ratios

Some trigonometric solutions based problems on trigonometric ratios are shown here with the step-by-step explanation.

1. If sin θ = 8/17, find other trigonometric ratios of <θ.

Let us draw a ∆ OMP in which ∠M = 90°.

Then sin θ = MP/OP = 8/17.

Let MP = 8k and OP = 17k, where k is positive.

By Pythagoras’ theorem, we get

Therefore, sin θ = MP/OP = 8k/17k = 8/17

cos θ = OM/OP = 15k/17k = 15/17

tan θ = Sin θ/Cos θ = (8/17 × 17/15) = 8/15

csc θ = 1/sin θ = 17/8

sec θ = 1/cos θ = 17/15 and

cot θ = 1/tan θ = 15/8.

2. If Cos A = 9/41, find other trigonometric ratios of ∠A.

Let us draw a ∆ ABC in which ∠B = 90°.

Then cos θ = AB/AC = 9/41.

Let AB = 9k and AC = 41k, where k is positive.

Therefore, sin A = BC/AC = 40k/41k = 40/41

cos A = AB/AC = = 9k/41k = 9/41

tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9

csc A = 1/sin A = 41/40

sec A = 1/cos A = 41/9 and

cot A = 1/tan A = 9/40.

3. Show that the value of sin θ and cos θ cannot be more than 1.

We know, in a right angle triangle the hypotenuse is the longest side.

sin θ = perpendicular/hypotenuse = MP/OP < 1 since perpendicular cannot be greater than hypotenuse; sin θ cannot be more than 1.

Similarly, cos θ = base/hypotenuse = OM/OP < 1 since base cannot be greater than hypotenuse; cos θ cannot be more than 1.

4. Is that possible when A and B be acute angles, sin A = 0.3 and cos B = 0.7?

Since A and B are acute angles, 0 ≤ sin A ≤ 1 and 0 ≤ cos B ≤ 1, that means the value of sin A and cos B lies between 0 to 1. So, it is possible that sin A = 0.3 and cos B = 0.7

5. If 0° ≤ A ≤ 90° can sin A = 0.4 and cos A = 0.5 be possible?

Let us draw a ∆ ABC in which ∠B = 90° and ∠BAC = θ.

Then sin θ = BC/AC = 1/2.

Let BC = k and AC = 2k, where k is positive.

= 3√3/2 - 4 × 3√3/8

= 3√3/2 - 3√3/2

Hence, (3cos θ - 4 cos³ θ) = 0.

7. Show that sin α + cos α > 1 when 0 ° ≤ α ≤ 90°

From the right triangle MOP,

Sin α = perpendicular/ hypotenuse

Cos α = base/ hypotenuse

Now, Sin α + Cos α

= perpendicular/ hypotenuse + base/ hypotenuse

= (perpendicular + base)/hypotenuse, which is > 1, Since we know that the sum of two sides of a triangle is always greater than the third side.

8. If cos θ = 3/5, find the value of (5csc θ - 4 tan θ)/(sec θ + cot θ)

Let ∠A = θ°

Then cos θ = AB/AC = 3/5.

Let AB = 3k and AC = 5k, where k is positive.

Therefore, sec θ = 1/cos θ = 5/3

tan θ = BC/AB =4k/3k = 4/3

cot θ = 1/tan θ = 3/4 and

csc θ = AC/BC = 5k/4k = 5/4

Now (5csc θ -4 tan θ)/(sec θ + cot θ)

= (5 × 5/4 - 4 × 4/3)/(5/3 + 3/4)

= (25/4 -16/3)/(5/3 +3/4)

= 11/12 × 12/29

9. Express 1 + 2 sin A cos A as a perfect square.

1 + 2 sin A cos A

10. If sin A + cos A = 7/5 and sin A cos A =12/25, find the values of sin A and cos A.

sin A + cos A = 7/5

⇒ cos A = 7/5 - sin θ

Now from sin θ/cos θ = 12/25

We get, sin θ(7/5 - sin θ) = 12/25

or, 5 sin θ(5 sin θ - 4) - 3(5 sin θ - 4) = 0

or, (5 sin θ - 3) (5 sin θ - 4) = 0

⇒ (5 sin θ - 3) = 0 or, (5 sin θ - 4) = 0

⇒ sin θ = 3/5 or, sin θ = 4/5

When sin θ = 3/5, cos θ = 12/25 × 5/3 = 4/5

Again, when sin θ = 4/5, cos θ = 12/25 × 5/4 = 3/5

Therefore, sin θ =3/5, cos θ = 4/5

or, sin θ =4/5, cos θ = 3/5.

11. If 3 tan θ = 4, evaluate (3sin θ + 2 cos θ)/(3sin θ - 2cos θ).

Solution: Given,

3 tan θ = 4

⇒ tan θ = 4/3

(3sin θ + 2 cos θ)/(3sin θ - 2cos θ)

= (3 tan θ + 2)/(3 tan θ - 2), [dividing both numerator and denominator by cos θ]

= (3 × 4/3 + 2)/(3 × 4/3 -2), putting the value of tan θ = 4/3

12. If (sec θ + tan θ)/(sec θ - tan θ) = 209/79, find the value of θ.

Solution: (sec θ + tan θ)/(sec θ - tan θ) = 209/79

⇒ [(sec θ + tan θ) - (sec θ - tan θ)]/[(sec θ + tan θ) + (sec θ - tan θ)] = [209 – 79]/[209 + 79], (Applying componendo and dividendo)

⇒ 2 tan θ/2 sec θ =130/288

⇒ sin θ/cos θ × cos θ = 65/144

⇒ sin θ = 65/144.

13. If 5 cot θ = 3, find the value of (5 sin θ - 3 cos θ)/(4 sin θ + 3 cos θ).

Given 5 cot θ = 3

⇒ cot θ = 3/5

Now (5 sin θ - 3 cos θ)/(4 sin θ + 3 cos θ)

= (5 - 3 cot θ)/(4 sin θ + 3 cot θ), [dividing both numerator and denominator by sin θ]

= (5 - 3 × 3/5)/(4 + 3 × 3/5)

= (5 - 9/5)/(4 + 9/5)

= (16/5 × 5/29)

⇒ sin θ(sin θ - 2) - 1(sin θ - 2) = 0

⇒ (sin θ - 2)(sin θ - 1) = 0

⇒ (sin θ - 2) = 0 or, (sin θ - 1) = 0

⇒ sin θ = 2 or, sin θ = 1

So, value of sin θ can’t be greater than 1,

Therefore sin θ = 1

Basic Trigonometric Ratios  

Relations Between the Trigonometric Ratios

Reciprocal Relations of Trigonometric Ratios

Trigonometrical Identity

Problems on Trigonometric Identities

Elimination of Trigonometric Ratios  

Eliminate Theta between the equations

Problems on Eliminate Theta  

Trig Ratio Problems

Proving Trigonometric Ratios

Trig Ratios Proving Problems

Verify Trigonometric Identities  

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Solve Problems Using Trigonometric Ratios

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Trigonometry : Solving Word Problems with Trigonometry

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry.

You can draw the following right triangle using the information given by the question:

Since you want to find the height of the platform, you will need to use tangent.

You can draw the following right triangle from the information given by the question.

In order to find the height of the flagpole, you will need to use tangent.

You can draw the following right triangle from the information given in the question:

In order to find out how far up the ladder goes, you will need to use sine.

In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC?

This triangle cannot exist.

Example Question #5 : Solving Word Problems With Trigonometry

A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.

23.81 meters

28.31 meters

21.83 meters

To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. 

Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.

We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore:

To solve this problem instead using the cosecant function, we would get:

The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem. 

Example Question #6 : Solving Word Problems With Trigonometry

When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?

To solve this problem, first set up a diagram that shows all of the info given in the problem. 

Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b.

Therefore the shadow cast by the building is 150 meters long.

If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)

Example Question #7 : Solving Word Problems With Trigonometry

From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?

423.18 meters

318.18 meters

36.15 meters

110.53 meters

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.

Merging together the given info and this diagram, we know that the angle of depression is 19 o  and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get:

Example Question #8 : Solving Word Problems With Trigonometry

Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?

9.37 meters

1480 meters

3708.74 meters

10677.87 meters

1616.1 meters

As with other trig problems, begin with a sketch of a diagram of the given and sought after information.

Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:

Therefore the change in height between Angelina's starting and ending points is 1480 meters. 

Example Question #9 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?

To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle. 

Example Question #10 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building?

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Chapter 5: Equations and Identities

5.2 Solving Equations

Algebra refresher.

Each of the following “solutions” contains an error. Find the error and supply a correct solution.

• [latex]\displaystyle [latex] \begin{aligned}[t] 3x^2 - 5x = 0 \\ 3x^2 = 5x\\ 3x = 5\\ x = \dfrac{5}{3}\\ \end{aligned}[/latex]
• [latex]\displaystyle [latex] \begin{aligned}[t] 4x^2 - 5x =12 \\ x^2 = 3\\ x = \\\sqrt{3}\\ \end{aligned}[/latex]
• [latex]\displaystyle [latex] \begin{aligned}[t] (2x+1)^2 = 4 \\ 2x+1 = 2\\ 2x = 1\\ x = \dfrac{1}{2}\\ \end{aligned}[/latex]
• [latex]\displaystyle [latex] \begin{aligned}[t] 2x^2 - 4x = 3 \\ 2x(x-2) = 3\\ 2x = 3 \qquad x - 2 = 3\\ x = \dfrac{3}{2} \qquad x = 5\\ \end{aligned}[/latex]

[latex]\underline{\qquad\qquad\qquad\qquad}[/latex]

• We can't divide by [latex]x.~~x = 0, \dfrac{5}{3}[/latex]
• 3 has two square roots. [latex]~x = \pm\sqrt{3}[/latex]
• 4 has two square roots. [latex]~x = \dfrac{-3}{2}, \dfrac{1}{2}[/latex]
• One side must be zero to use the zero-factor principle. No real solutions.

Learning Objectives

• Use reference angles
• Solve equations by trial and error
• Use graphs to solve equations
• Solve trigonometric equations for exact values
• Use a calculator to solve trigonometric equations
• Solve trigonometric equations that involve factoring

Introduction

It is important to distinguish between an algebraic expression and an equation . An equation is a statement that two algebraic expressions are equal. It may be true or false, depending on the values of any variables involved. Here are some examples of equations.

[latex]5(2 + 6) = 5(2) + 5(6)[/latex] [latex]\sqrt{3^2} + \sqrt{4^2}= 3 + 4[/latex] [latex]x^2 + 3x = 10[/latex]

The first equation is true, the second is false, and the third equation is true only if [latex]x = 2[/latex] or [latex]x = -5{.}[/latex] When you solve an equation, you are finding the values of the variable that make the equation true.

Example 5.19

• Evaluate [latex]2x + \sqrt{x - 5}[/latex] for [latex]x = 9{.}[/latex]
• Solve [latex]2x + \sqrt{x - 5} = 20{.}[/latex]
• Substitute [latex]x = 9[/latex] into the expression to find [latex]2x + \sqrt{x - 5} = 2({9}) + \sqrt{{9} - 5}= 18 + \sqrt{4} = 20[/latex]
• We must find a value for [latex]x[/latex] that makes [latex]2x + \sqrt{x - 5}[/latex] equal to 20. In part (a) we saw that this expression equals 20 when [latex]x = 9{,}[/latex] so the solution of the equation [latex]2x + \sqrt{x - 5} = 20[/latex] is [latex]x = 9{.}[/latex]

Checkpoint 5.20.

Try small integer values for [latex]x{.}[/latex]

[latex]x = 3[/latex]

You probably remember a number of algebraic techniques for solving equations of different types. Another useful equation-solving method uses graphs.

Example 5.21.

Use a graph to solve the equation [latex]x^3 - 2x^2 - 5x = -6{.}[/latex]

Checkpoint 5.22.

The graph does not cross the line [latex]y = 0{.}[/latex]

Trigonometric Equations

The first Ferris wheel was built for the Chicago World's Fair in 1893. It had a diameter of 250 feet and could carry 2160 people in 36 carriages. From the top of the wheel, passengers could see into four states. After loading all the passengers, the wheel made one revolution in nine minutes.

If you are in the bottom carriage of the Ferris wheel at the start of its revolution, your height after [latex]t[/latex] seconds is given by

[latex]h = f(t) = 139 - 125\cos \left(\dfrac{2t}{3}\right)[/latex]

For how long are you more than 240 feet above the ground?

The figure below shows a graph of the height function and a horizontal line at [latex]h = 240{.}[/latex]

Solving Trigonometric Equations

In the example above, we used a graph to solve the equation [latex]h = 240{,}[/latex] or

[latex]139 - 125\cos \left(\dfrac{2t}{3}\right) = 240[/latex]

To find a more precise solution, we can use algebraic methods. As an example, we'll solve the slightly simpler equation

[latex]139 - 125\cos\theta = 240[/latex]

We'll look for all solutions for [latex]\theta[/latex] between [latex]0°[/latex] and [latex]360°{.}[/latex] We begin by isolating the trigonometric ratio on one side of the equation.

[latex]139 - 125\cos \theta = 240[/latex] Subtract 139 from both sides.} [latex]- 125\cos \theta = 101[/latex] Divide both sides by -125. [latex]\cos\theta = -0.808[/latex]

We have solved equations like this one before: we use the inverse cosine to solve for [latex]\theta{.}[/latex] Remember that there are two angles between [latex]0°[/latex] and [latex]360°[/latex] that have a cosine of [latex]-0.808{,}[/latex] one in the second quadrant and one in the third quadrant. The calculator will give us only the second quadrant solution.

[latex]\theta = \cos^{-1}(-0.808) = 143.9°[/latex]

To find the second solution, we need the third-quadrant angle whose cosine is [latex]-0.808{.}[/latex]

Now, the reference angle for [latex]143.9°[/latex] is

[latex]180° - 143.9° = 36.1°[/latex]

and the angle in the third quadrant with the same reference angle is

[latex]180° + 36.1° = 216.1°[/latex]

(See the figure at right.) Thus, the other solution is [latex]216.1°{.}[/latex]

To solve simple equations involving a single trigonometric ratio (either [latex]\sin \theta, \cos \theta,[/latex] or [latex]\tan \theta[/latex]), we can follow the steps below.

To Solve a Trigonometric Equation

1. Isolate the trigonometric ratio.

2. Find one solution in [latex]0° \le\theta\le 360°[/latex].

a. Give an exact solution if the trig ratio is one of the special values.

b. Otherwise use the inverse trig keys on a calculator.

3. Use reference angles to find a second solution (if there is one).

In step 3, note that if the first solution is a quadrantal angle, there may not be a second solution. Consider, for example, the equation [latex]\cos \theta = -1{.}[/latex]

Example 5.24.

Solve the equation [latex]~~8 \sin \theta - 1 = 3~~[/latex] for [latex]0° \le\theta\le 360°{.}[/latex]

We begin by isolating the trig ratio.

[latex]8 \sin \theta - 1 = 3[/latex] Add 1 to both sides. [latex]8 \sin \theta = 4[/latex] Divide both sides by 8. [latex]\sin \theta = \dfrac{4}{8} = \dfrac{1}{2}[/latex]

Caution 5.25.

To solve the equation in the previous example, it is not enough to find the solution [latex]\theta = 30°{;}[/latex] we must find all the solutions between [latex]0°[/latex] and [latex]360°{.}[/latex] To find the solutions in other quadrants, we use reference angles.

Each intersection represents a solution of the equation. Thus, all of the angles coterminal with [latex]30°[/latex] and [latex]150°[/latex] are also solutions. We can easily find these solutions by adding integer multiples of [latex]360°[/latex] to [latex]30°[/latex] or [latex]150°{.}[/latex] This is why, when solving a trigonometric equation, we usually list only the solutions in one cycle, typically those between [latex]0°[/latex] and [latex]360°{.}[/latex]

Checkpoint 5.27.

[latex]\theta = 135°~ {and}~ \theta = 315°[/latex]

We can use a calculator to help us solve equations that do not involve special angles.

Example 5.28.

First, we isolate the trig ratio.

[latex]3 \tan \beta + 1 = -8\\ 3 \tan \beta = -9\\ \tan \beta = -3[/latex]

There are two angles with tangent -3, one in the second quadrant and one in the fourth quadrant. The calculator finds the solution

[latex]\tan^{-1}(-3) = -71.565°[/latex]

This angle is in the fourth quadrant, but it is not between [latex]0°[/latex] and [latex]360°{.}[/latex]

The angle we want is coterminal with [latex]-71.565°{,}[/latex] so we add [latex]360°[/latex] to get the first solution:

[latex]B = \tan^{-1}(-3) + 360° = 288.435°[/latex]

The other solution is in the second quadrant, as shown in the figure. This angle is [latex]180°[/latex] less than the fourth quadrant solution:

[latex]B = \tan^{-1}(-3) + 180° = 108.435°[/latex]

In the previous example, notice that the solutions are [latex]180°[/latex] apart. The solutions of an equation [latex]\tan \theta = k[/latex] always differ by [latex]180°{,}[/latex] so the use of reference angles is not necessary for these equations: once we have found one solution, we can add or subtract [latex]180°[/latex] to find the other solution between [latex]0°[/latex] and [latex]360°{.}[/latex] (See Homework Problems 69 and 70.)

Checkpoint 5.30.

[latex]C = 41.410°~[/latex] or [latex]~ C = 318.590°[/latex]

Some trigonometric equations have no solution. As we can observe from their graphs or from their definitions, the sine and cosine functions only have values ranging from [latex]-1[/latex] to [latex]1{.}[/latex]

Example 5.31.

Solve [latex]~~\dfrac{\sin \beta}{5} - 3 = 1~~[/latex] for [latex]0° \le\theta\le 360°{.}[/latex]

We begin by isolating [latex]\sin \beta{.}[/latex]

[latex]\dfrac{\sin \beta}{5} - 3 = 1[/latex] Add 3 to both sides. [latex]\dfrac{\sin \beta}{5} = 4[/latex] Multiply both sides by 5. [latex]\sin \beta = 20[/latex]

Because [latex]\sin \beta[/latex] is never greater than 1, there is no angle [latex]\beta[/latex] whose sine is 20. The equation has no solution.

Checkpoint 5.32.

No solution

Graphical Solutions

We can use graphs to approximate the solutions to trigonometric equations.

Example 5.33.

Use a graph to solve the equation [latex]~~3\tan B + 1 = -8{.}[/latex]

Checkpoint 5.34.

Equations with Squares of Trig Ratios

Simple quadratic equations can be solved by extracting roots. For example, to solve the equation

[latex]4x^2 + 3 = 15[/latex]

we first isolate [latex]x^2{:}[/latex]

[latex]4x^2 = 12\\ x^2 = 3[/latex]

and then take square roots of both sides to find

[latex]x = \pm \\\sqrt{3} \approx \pm 1.732[/latex]

Recall that a quadratic equation may have two real solutions, one (repeated) real solution, or no real solutions. We can use extraction of roots to solve trigonometric equations as well.

Example 5.35.

We begin by isolating the power of tangent, [latex]\tan^2 \theta{.}[/latex]

[latex]4\tan^2 \theta = 12\\ \tan^2 \theta = 3[/latex]

Next, we solve for [latex]\tan \theta[/latex] by extracting roots.

[latex]\tan \theta = \pm \\\sqrt{3}[/latex]

There are two angles between [latex]0°[/latex] and [latex]360°[/latex] with tangent [latex]\\\sqrt{3}[/latex] and two angles with tangent [latex]-\\\sqrt{3}{,}[/latex] making four solutions to this equation. We know that [latex]\tan 60° = \\\sqrt{3}{,}[/latex] so one of the solutions is [latex]60°{.}[/latex]

[latex]180° - 60° = 120° ~~ {and} ~~ 360° - 60° = 300°[/latex]

The four solutions are shown at right.

Checkpoint 5.36.

[latex]\theta = 45°{,}[/latex] [latex]~\theta = 135°{,}[/latex] [latex]~\theta = 225°~{,}[/latex] or [latex]~\theta = 315°[/latex]

Other quadratic equations can be solved by factoring. For example, we can solve the equation

[latex]4x^2 + 4x - 3 = 0[/latex]

by factoring the left side to get

[latex](2x + 3)(2x - 1) = 0[/latex]

Then we apply the Zero Factor Principle to set each factor equal to zero, and solve each equation.

[latex]2x + 3 = 12 2x - 1= 0\\ x = \dfrac{-3}{2} x = \dfrac{1}{2}[/latex]

The solutions are [latex]\dfrac{-3}{2}[/latex] and [latex]\dfrac{1}{2}{.}[/latex]

Example 5.37.

Compare this equation to the algebraic equation [latex]2x^2 - x - 1 = 0{.}[/latex] Our equation has the same algebraic form, but with [latex]x[/latex] replaced by [latex]\cos \theta {.}[/latex]

We can solve the algebraic equation by factoring [latex]2x^2 - x - 1[/latex] as [latex](2x + 1)(x - 1),[/latex] and we'll use the same strategy on the trigonometric equation.

[latex]2 \cos^2 \theta - \cos \theta - 1 = 0[/latex] Factor the left side. [latex](2\cos \theta + 1)(\cos \theta - 1) = 0[/latex] Set each factor equal to zero. [latex]\cos \theta = \dfrac{-1}{2} ~~ { or } ~~\cos \theta = 1[/latex]

Now we solve each equation for [latex]\theta{.}[/latex] We know that [latex]\cos 60° = \dfrac {1}{2},[/latex] and the cosine function is negative in the second and third quadrants. The angles in those quadrants with reference angle [latex]60°[/latex] are

[latex]180° - 60° = 120° ~~~{and} ~~~ 180° + 60° = 240°[/latex]

Also, [latex]\cos \theta = 1[/latex] when [latex]\theta = 0°{.}[/latex] Thus, the original equation has three solutions: [latex]\theta = 120°, ~ \theta = 240°,[/latex] and [latex]\theta = 0°.[/latex]

  Checkpoint 5.38.

[latex]\theta = 42°{,}[/latex] [latex]~\theta = 138°{,}[/latex] [latex]~\theta = 199°~{,}[/latex] or [latex]~\theta = 341°[/latex]

Snell's Law

When you view an object through a liquid, such as a spoon in a glass of water, or a fish in an aquarium, the object looks distorted or bent. This distortion is caused by refraction of light. Light rays bend when they pass from one medium to another, for instance from water to glass or from glass to air.

A light ray enters the boundary between the two media at a certain angle, called the angle of incidence , but leaves the boundary at a different angle, the angle of refraction . Both angles are acute angles measured from the normal line perpendicular to the boundary, as shown below.

The change of angle is caused by the fact that light travels at different speeds in different media. The relationship between the angle of incidence and the angle of refraction is given by Snell's Law:

[latex]\dfrac{\sin \theta_1}{\sin \theta_2} = \dfrac{v_1}{v_2}[/latex]

where [latex]\theta_1[/latex] is the angle in the medium where light travels at speed [latex]v_1{,}[/latex] and [latex]\theta_2[/latex] is the angle where light travels at speed [latex]v_2{.}[/latex] The ratio of the speeds is called the index of refraction .

Example 5.39.

The index of refraction from water to glass is 1.1. If light passes from water to glass with a [latex]23°[/latex] angle of incidence, what is the angle of refraction?

If [latex]\theta[/latex] is the angle of refraction, then from Snell's Law we have

[latex]\dfrac{\sin 23°}{\sin \theta} = 1.1[/latex] Multiply both sides by} \sin \theta. [latex]\sin 23° = 1.1 \sin \theta[/latex] Divide both sides by 1.1. [latex]\sin \theta = \dfrac{\sin 23°}{1.1} = 0.3552[/latex]

Because [latex]\sin \theta = 0.3552{,}[/latex] [latex]~ \theta = \sin^{-1}(0.3552) = 20.8°.[/latex] (For Snell's Law we use only acute angles.) The angle of refraction is approximately [latex]20.8°{.}[/latex]

Checkpoint 5.40.

A light ray passes from water to glass with an [latex]18°[/latex] angle of incidence. What is the angle of refraction?

[latex]16.315°[/latex]

Section 5.2 Summary

• Zero Factor Principle
• Angle of incidence
• Angle of refraction

                                          Concepts

• An equation is a statement that two algebraic expressions are equal. It may be true or false.
• We can solve equations by trial and error, by using graphs, or by algebraic techniques.
• To solve a trigonometric equation, we first isolate the trigonometric ratio on one side of the equation.
• We use reference angles to find all the solutions between [latex]0°[/latex] and [latex]360°{.}[/latex]
• We can use factoring or extraction of roots to solve some quadratic equations.

                                    Study Questions

• How many solutions between [latex]0°[/latex] and [latex]360°[/latex] does the equation [latex]\cos \theta = k[/latex] have for each value of [latex]k[/latex] between -1 and 0?
• How many solutions between [latex]0°[/latex] and [latex]360°[/latex] does the equation [latex]\cos \theta = k[/latex] have for each value of [latex]k[/latex] greater than 1?
• How many solutions between [latex]0°[/latex] and [latex]360°[/latex] does the equation [latex]\sin^2 \theta = k[/latex] have for each value of [latex]k[/latex] between -1 and 0?

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Trigonometric Ratios

A series of free, online High School Geometry Video Lessons and solutions. In these lessons, we will learn

• how to find the sine of an angle in a right triangle,
• how to find the cosine of an angle in a right triangle,
• how to find the tangent of an angle in a right triangle,
• how to use inverse trigonometric functions to find an angle with a given trigonometric value.

Related Pages Using SOH-CAH-TOA Trigonometry Word Problems Inverse trigonometry Lessons On Trigonometry Trigonometry Worksheets

The following diagram shows the trigonometric ratios using SOHCAHTOA. Scroll down the page if you need more examples and solutions on how to use the trigonometric ratios.

Trigonometric Ratios: Sine

Right triangles have ratios to represent the angles formed by the hypotenuse and its legs. Sine ratios, along with cosine and tangent ratios, are ratios of the lengths of two sides of the triangle. Sine ratios in particular are the ratios of the length of the side opposite the angle they represent over the hypotenuse. Sine ratios are useful in trigonometry when dealing with triangles and circles.

How to define the sine ratio and identify the sine of an angle in a right triangle?

Identify the hypotenuse of a right triangle. sin θ = opposite/hypotenuse

A word problem involving the trigonometric ratio of sine to calculate the height of a pole

Example: A 55 ft wire connects a point on the ground to the top of a pole. The cable makes an angle of 60° with the ground. Find the height of the pole to the nearest foot.

Trigonometric Ratios: Cosine

Right triangles have ratios that are used to represent their base angles. Cosine ratios, along with sine and tangent ratios, are ratios of two different sides of a right triangle.Cosine ratios are specifically the ratio of the side adjacent to the represented base angle over the hypotenuse. In order to find the measure of the angle, we must understand inverse trigonometric functions.

How to use the Cosine formula (the CAH Formula) to find missing sides or angles?

Cosine θ = adjacent/hypotenuse

Example: Find the missing side or angle a) cos 30° = x/2 b) cos 37° = 4.2/x c) cos θ = 63/80

How to apply the Sine and Cosine Ratios?

Sine, cosine are trigonometric ratios for the acute angles and involve the length of a leg and the hypotenuse of a right triangle.

Angle of elevation - When looking up at an object, the angle your line of sight makes with a horizontal line is called the angle of elevation.

Angle of depression - When looking down at an object, the angle your line of sight makes with a horizontal line is called the angle of depression.

• Find sin U and sin W. Write each answer as a fraction and as a decimal rounded to four places.
• Find cos S and cos R. Write each answer as a fraction and as a decimal rounded to four places.
• You walk from one corner of a basketball court to the opposite corner. write and solve a proportion using a trigonometric ratio to approximate the distance of the walk.
• You are at the top of a roller coaster 100 feet above the ground. The angle of depression is 44°. About how far do you ride down the hill?
• A railroad crossing arm that is 20 feet long is stuck with an angle of elevation of 35°. Find the lengths of x and y.
• Use a special right triangle to find the sine and cosine of a 30° angle.

Trigonometric Ratios: Tangent

Right triangles have ratios that are used to represent their base angles. Tangent ratios, along with cosine and sine ratios, are ratios of two different sides of a right triangle. Tangent ratios are the ratio of the side opposite to the side adjacent the angle they represent. In order to find the measure of the angle itself, one must understand inverse trigonometric functions.

How to use the Tangent formula (the TOA Formula)?

Tangent θ = opposite/adjacent

Example: Find the missing side or angle a) tan 28° = x/40 b) tan 41° = 1.9/x c) tan θ = 11/8

Applications of Trigonometric Ratios (Word Problems Involving Tangent, Sine and Cosine)

• Find the area of the parallelogram.
• A 70 foot ramp rises from the first floor to the second floor of a parking garage. The ramp makes an angle with the ground. How high above the first floor is the second floor?
• You see Mr. Wandera flying a kite in the park. The kite string is 65 meters long. What angle does the string need to form with the ground so that the kite is 30 feet off of the ground?
• From the top of a 100-foot lookout tower, a forest ranger spotted a fire at a 25° angle of depression. How far was the fire from the base of the lookout tower?
• An 8 foot ladder is leaning against a wall. The ladder makes a 53° angle with the wall. How high does the ladder reach?

Inverse Trigonometric Functions

Once we understand the trigonometric functions sine, cosine, and tangent, we are ready to learn how to use inverse trigonometric functions to find the measure of the angle the function represents. Inverse trigonometric functions, found on any standard scientific or graphing calculator, are a vital part of trigonometry and will be encountered often in Calculus.

How to use inverse trigonometric functions to find an angle with a given trigonometric value and how to use inverse trigonometric functions to solve a right triangle?

Example: Use the calculator to find an angle θ in the interval [0, 90] that satisfies the equation.

• sin θ = 0.7523
• tan θ = 3.54
• Solve the given right triangle if a = 44.3 cm and b = 55.9 cm.
• Find each angle in a 3,4,5 triangle

How to use inverse trig to find a missing angle?

Inverse trig functions are used to find missing angles rather than missing sides.

Find missing Angles - Using Inverse Sine, Cosine, Tangent

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Examples for

Step-by-Step Trigonometry

Use Wolfram|Alpha to get information about trigonometry problems, from solving for missing measurements of triangles to evaluating trigonometric functions to manipulating trigonometric expressions. See step-by-step solutions for using trigonometric ratios like sine, cosine and tangent to compute the measurements of sides and angles of triangles. Learn how to find values of additional trigonometric functions, including secant, cosecant and cotangent. Get explanations for how to algebraically manipulate expressions involving trigonometric functions, including to prove and apply identities such as the Pythagorean identity, the negative-angle identity, and the sum and difference identities.

Right Angle Trigonometry

Use trigonometric ratios to compute measurements of right triangles.

Learn how to find the side lengths of a right triangle using the Pythagorean theorem:

Show your work for solving the unknown measurements of a triangle:, trigonometric expressions.

Solve and manipulate expressions involving trigonometric functions.

Explore the steps for solving trigonometric equations:

Expand trigonometric expressions with the steps provided:, perform double angle reductions:, show the steps to expand expressions using the triple angle formulas:, trigonometric functions.

Evaluate sine, cosine, tangent, secant, cosecant and cotangent functions.

See the steps for how to find the values of a trigonometric function:

Trigonometric identities.

Prove and reason with trigonometric identities.

See the steps for how to apply the Pythagorean identity:

Show how to apply the negative-angle identity using even and odd properties step by step:, get step-by-step solutions for applying the sum and difference formulas:, explain step by step how to prove trigonometric identities:, related examples.

• Oscillations & Waves
• Periodic Functions
• Trigonometry

• Math Article

Trigonometric Ratios

The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec). In geometry, trigonometry is a branch of mathematics that deals with the sides and angles of a right-angled triangle. Therefore, trig ratios are evaluated with respect to sides and angles. 

The trigonometry ratios for a specific angle ‘θ’ is given below:

Note: Opposite side is the perpendicular side and the adjacent side is the base of the right-triangle. Also, check out trigonometric functions to learn about each of these ratios or functions in detail. Trigonometric Identities

Trigonometric Ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.   The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle .

The three sides of the right triangle are:

• Hypotenuse (the longest side)
• Perpendicular (opposite side to the angle)
• Base (Adjacent side to the angle)

Related Articles:

• Trigonometric Identities
• Trigonometric Ratios of Complementary Angles
• Trigonometric Ratios Of Standard Angles
• Trigonometry Angles
• Trigonometry Formulas
• Trigonometry Values

How to Find Trigonometric Ratios?

Consider a right-angled triangle, right-angled at B.

With respect to ∠C, the ratios of trigonometry are given as:

• sine: Sine of an angle is defined as the ratio of the side opposite(perpendicular side) to that angle to the hypotenuse.
• cosine: Cosine of an angle is defined as the ratio of the side adjacent to that angle to the hypotenuse.
• tangent: Tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle.
• cosecant: Cosecant is a multiplicative inverse of sine.
• secant: Secant is a multiplicative inverse of cosine.
• cotangent: Cotangent is the multiplicative inverse of the tangent.

The above ratios are abbreviated as sin, cos, tan, cosec, sec and tan respectively in the order they are described. So, for Δ ABC , the ratios are defined as:

sin C = ( Side opposite to ∠C)/(Hypotenuse)  = AB/AC

cos C  = ( Side adjacent to ∠C)/(Hypotenuse)  = BC/AC

tan C  = ( Side opposite to ∠C)/(Side adjacent to ∠C)  = AB/BC  = sin ∠C/cos ∠C

cosec C = 1/sin C = ( Hypotenuse)/ (Side Opposite to ∠C)  =  AC/AB

sec C  = 1/cos C = ( Hypotenuse)/ (Side Opposite to ∠C)  = AC/BC

cot C  = 1/tan C  = ( Side adjacent to ∠C)/(Side opposite to ∠C) = BC/AB

In right Δ ABC,  if ∠ A  and ∠ C  are assumed as 30°  and 60° , then there can be infinite right triangles with those specifications but all the ratios written above for ∠ C  in all of those triangles will be same. So, all the ratios for any of the acute angles (either ∠ A  or ∠ C ) will be the same for every right triangle. This means that the ratios are independent of lengths of sides of the triangle.

Trigonometric Ratios Table

The trigonometric ratios for some specific angles such as 0 °, 30 °, 45 °, 60 ° and 90° are given below, which are commonly used in mathematical calculations.

From this table, we can find the value for the trigonometric ratios for these angles. Examples are:

• Sin 30° = ½
• Cos 90° = 0
• Tan 45° = 1 

Trigonometry Applications

Trigonometry is one of the most important branches of mathematics. Some of the applications of trigonometry are:

• Measuring the heights of towers or big mountains
• Determining the distance of the shore from the sea
• Finding the distance between two celestial bodies
• Determining the power output of solar cell panels at different inclinations
• Representing different physical quantities such as mechanical waves, electromagnetic waves, etc.

It is evident from the above examples that trigonometry has its involvement in a major part of our day-to-day life and much more. In most of the applications listed above, something was being measured and that is what trigonometry is all about.

Solved Problems

Q.1: If in a right-angled triangle ABC, right-angled at B, hypotenuse AC = 5cm, base BC = 3cm and perpendicular AB = 4cm and if ∠ACB = θ, then find tan θ, sin θ and cos θ.

Sol: Given, 

Hypotenuse, AC = 5cm 

Base, BC = 3cm 

Perpendicular, AB = 4cm

tan θ = Perpendicular/Base = 4/3

Sin θ = Perpendicular/Hypotenuse = AB/AC = ⅘

Cos θ = Base/Hypotenuse = BC/AC = ⅗

Q.2: Find the value of tan θ if sin θ = 12/5 and cos θ = ⅗.

Sol: Given, sin θ = 12/5 and cos θ = ⅗

As we know,

Tan θ = Sin θ/Cos θ

Tan θ = (12/5)/(⅗)

Tan θ = 12/3

Practice Questions

• Find the value of sin θ, if tan θ = ¾ and cos θ = ½. 
• Find tan θ if sin θ = 4/3 and cos θ = 3/2
• Find sec θ, if cos θ = 9/8
• Find cosec θ, if sin θ 16/5

Video Lesson

Trigonometric ratios of compound angles.

Frequently Asked Questions – FAQs

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TRIGONOMETRIC RATIOS EXAMPLES AND SOLUTIONS

Example problems and solutions given in this section will be much useful for the students who would like to practice problems on trigonometric ratios.

Before look at the example problems, if you would like to know the basic stuff on trigonometric ratios,

please click here

Example 1 :

Compare the sine, the cosine, and the tangent ratios for  ∠ A in each triangle below.

By the SSS Similarity Theorem, the triangles are similar. Their corresponding sides are in proportion, which implies that the trigonometric ratios for  ∠ A in each triangle are the same.

Trigonometric ratios are frequently expressed as decimal approximations.

Example 2 :

Find the sine, the cosine, and the tangent of the indicated angle.

a.   ∠S          b.   ∠R

Solution (a) :

The length of the hypotenuse is 13. For  ∠ S, the length of the opposite side is 5, and the length of the adjacent side is 12.

sin S  =  opp./hyp.  =  5/13   ≈  0.3846

cos S  =  adj./hyp.  =  12/13   ≈  0.9231

tan S  =  opp./adj.  =  5/12   ≈  0.4167

Solution (b) :

The length of the hypotenuse is 13. For  ∠ R, the length of the opposite side is 12, and the length of the adjacent side is 5.

sin R  =  opp./hyp.  =  12/13   ≈  0.9231

cos R  =  adj./hyp.  =  5/13   ≈  0.3846

tan R  =  opp./adj.  =  12/5   ≈  2.4

Example 3 :

Find the sine, the cosine, and the tangent of 45 °. 

Begin by sketching a 45°-45°-90° triangle. Because all such triangles are similar, we can make calculations simple by choosing 1 as the length of each leg. From the Pythagorean Theorem, it follows that the length of the hypotenuse is  √2.

sin 45 °   =  opp./hyp.  =  1/ √2  =   √2/2    ≈  0.7071

cos 45 °   =  adj./hyp.  =  1/ √2  =   √2/2   ≈  0.7071

tan 45 °   =  opp./adj.  =  1/1  =  1

Example 4 :

Find the sine, the cosine, and the tangent of 30 °. 

Begin by sketching a 30°-60°-90° triangle. To make the calculations simple, we can choose 1 as the length of the shorter leg. From  30°-60°-90° Triangle theorem ,  it follows that the length of the longer leg is  √3 and the hypotenuse is 2.

sin 30 °   =  opp./hyp.  =  1/ 2  =   0.5

cos 30 °   =  adj./hyp.  =  √3/2   ≈  0.8660   

tan 30 °   =  opp./adj.  =  1/ √3  =   √3/3   ≈  0.5774

Example 5 :

Jacob is measuring the height of a Sitka spruce tree in North Carolina. He stands 45 feet from the base of the tree. He measures the angle of elevation from a point on the ground to the top of the tree to be 59°. How can he estimate the height of the tree ? 

To estimate the height of the tree, Jacob can write a trigonometric ratio that involves the height h and the known length of 45 feet.

In the above right triangle, for the angle 59°, h is opposite side and the side has length 45 ft is adjacent side. 

The trigonometric ratio that involves opposite side and adjacent side is tangent. 

Write ratio : 

tan 59°  =  opp./adj. 

Substitute. 

tan 59°  =  h/45

Multiply each side by 45. 

45  ⋅ tan 59 °  =  h

Use calculator or table to find the value of tan 59 ° and substitute. 

45  ⋅ 1.6643  =   h

74.9   ≈  h

So, the tree is about 75 feet tall. 

Example 6 :

The escalator at the University Metro Rail Station near  University of Miami in Coral Gables, Florida  rises  76 feet at a 30° angle as shown in the diagram below. Find the distance d a person travels on the escalator stairs. 

To find the distance d a person travels on the escalator stairs, we  can write a trigonometric ratio that involves the side has length d and the known length of 76 feet.

In the above right triangle, for the angle 30°, d is hypotenuse and the side has length 76 feet is opposite side. 

The trigonometric ratio that involves opposite side and hypotenuse is sin. 

sin 30°  =  opp./hyp.

sin 30°  =  76/d

Multiply each side by d. 

d  ⋅ sin 30 °  =  76

Substitute 0.5 for sin  30 °.

d  ⋅ 0.5  =  76   

Divide each side by 0.5

d  =   76/0.5

d  =  152

So, the person travels 152 feet on the escalator stairs. 

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5. Trigonometry

5.2 Solve Applications: Sine, Cosine and Tangent Ratios.

Learning Objectives

By the end of this section it is expected that you will be able to

• Find missing side of a right triangle using sine, cosine, or tangent ratios
• Find missing angle of a right triangle using sine, cosine, or tangent ratios
• Solve applications using right angle trigonometry

Sine, Cosine, and Tangent Ratios

We know that any right triangle has three sides and a right angle. The side opposite to the right angle is called the hypotenuse. The other two angles in a right triangle are acute angles (with a measure less than 90 degrees). One of those angles we call reference angle and we use θ (theta) to represent it.

The hypotenuse is always the longest side of a right triangle. The other two sides are called opposite side and adjacent side. The names of those sides depends on which of the two acute angles is being used as a reference angle.

In the right triangle each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

Label the sides of the triangle and find the hypotenuse, opposite, and adjacent.

We labeled  the sides with a lowercase letter to match the uppercase letter of the opposite vertex.

c is hypotenuse

a is opposite

b is adjacent

Label the sides of the triangle and find the hypotenuse, opposite and adjacent.

y is hypotenuse

z is opposite

x is adjacent

Trigonometric Ratios

Trigonometric ratios are the ratios of the sides in the right triangle. For any right triangle we can define three basic trigonometric ratios: sine, cosine, and tangent.

Let us refer to Figure 1 and define the three basic trigonometric ratios as:

Three Basic Trigonometric Ratios

Where θ is the measure of a reference angle measured in degrees.

Very often we use the abbreviations for sine, cosine, and tangent ratios.

Some people remember the definition of the trigonometric ratios as SOH CAH TOA.

For the given triangle find the sine, cosine and tangent ratio.

For the given triangle find the sine cosine and tangent ratio.

When calculating we will usually round the ratios to four decimal places and at the end our final answer to one decimal place unless stated otherwise.

For the given triangle find the sine, cosine and tangent ratios. If necessary round to four decimal places.

We have two possible reference angles: R an S.

Using the definitions, the trigonometric ratios for angle R are:

Using the definitions, the trigonometric ratios for angle S:

For the given triangle find the sine, cosine, and tangent ratios. If necessary round to four decimal places.

Now, let us use a scientific calculator to find the trigonometric ratios.  Can you find the sin, cos, and tan buttons on your calculator? To find the trigonometric ratios make sure your calculator is in Degree Mode.

Using a calculator find the trigonometric ratios. If necessary, round to 4 decimal places.

Make sure your calculator is in Degree Mode.

a)  Using a calculator find that sin 30° = 0.5

b)  Using a calculator find that cos 45° = 0.7071 Rounded to 4 decimal places.

c)  Using a calculator find that tan 60° = 1.7321 Rounded to 4 decimal places.

Find the trigonometric ratios. If necessary, round to 4 decimal places.

a)  sin 60°

b)  cos 30°

c)  tan 45°

a)   sin 60° = 0.8660

b)   cos 30° = 0.8660

c)   tan 45° = 1

Finding Missing Sides of a Right Triangle

In this section you will be using trigonometric ratios to solve right triangle problems. We will adapt our problem solving strategy for trigonometry applications. In addition, since those problems will involve the right triangle, it is helpful to draw it (if the drawing is not given) and label it with the given information.We will include this in the first step of the problem solving strategy for trigonometry applications.

HOW TO: Solve Trigonometry Applications

• Read the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.
• Identify what we are looking for.
• Label what we are looking for by choosing a variable to represent it.
• Find the required trigonometric ratio.
• Solve the ratio using good algebra techniques.
• Check the answer by substituting it back into the ratio in step 4 and by making sure it makes sense in the context of the problem.
• Answer the question with a complete sentence.

In the next few examples, having given the measure of one acute angle and the length of one side of the right triangle, we will solve the right triangle for the missing sides.

 Find the missing sides. Round  your final answer to two decimal places

 Find the missing sides. Round  your final answer to one decimal place.

Find the hypotenuse. Round your final answer to one decimal place.

Finding Missing Angles of a Right Triangle

Sometimes we have a right triangle with only the sides given. How can we find the missing angles? To find the missing angles, we use the inverse of the trigonometric ratios. The inverse buttons sin -1 , cos -1 , and tan -1 are on your scientific calculator.

Find the angles. Round your final answer to one decimal place.

a)  sin A = 0.5

b)  cos B = 0.9735

c)  tan C = 2.89358

Use your calculator and press the 2nd FUNCTION key and then press the SIN, COS, or TAN key

a)  A = sin -1 0.5

b)  B = cos -1 0.9735

c) C = tan -1 2.89358

a)  sin X = 1

b)  cos Y = 0.375

c)  tan Z = 1.676767

In the example below we have a right triangle with two sides given. Our acute angles are missing. Let us see what the steps are to find the missing angles.

Find the missing angle X. Round your final answer to one decimal place.

Find the missing angle A. Round your final answer to one decimal place.

Find the missing angle C. Round your final answer to one decimal place.

Solving a Right Triangle

From the section before we know that any triangle has three sides and three interior angles. In a right triangle, when all six parts of the triangle are known, we say that the right triangle is solved.

Solve the right triangle. Round your final answer to one decimal place.

Since the sum of angles in any triangle is 180°, the measure of angle B can be easy calculated.

We solved the right triangle

Solve the right triangle. Round to two decimal places.

The missing angle F = 180° – 90° – 26.39° = 63.64°

Solve the right triangle. Round to one decimal place.

Solve Applications Using Trigonometric Ratios

In the previous examples we were able to find missing sides and missing angles of a right triangle. Now, let’s use the trigonometric ratios to solve real-life  problems.

Many applications of trigonometric ratios involve understanding of an angle of elevation or angle of depression.

The angle of elevation is an angle between the horizontal line (ground) and the observer’s line of sight.

The angle of depression is the angle between horizontal line (that is parallel to the ground) and the observer’s line of sight.

James is standing 31 metres away from the base of the Harbour Centre in Vancouver. He looks up to the top of the building at a 78° angle. How tall is the Harbour Centre?

Marta is standing 23 metres away from the base of the tallest apartment building in Prince George and looks at the top of the building at a 62° angle. How tall is the building?

43.3 metres

Thomas is standing at the top of the building that is 45 metres high and looks at her friend that is standing on the ground, 22 metres from the base of the building. What is the angle of depression?

Hemanth is standing on the top of a cliff 250 feet above the ground and looks at his friend that is standing on the ground, 40 feet from the base of the cliff. What is the angle of depression?

Key Concepts

• Read the problem and make sure all the words and ideas are understood. Draw the right triangle and label  the given parts.
• Check the answer by substituting it back into the ratio solved in step 5 and by making sure it makes sense in the context of the problem.

5.2 Exercise Set

Label the sides of the triangle.

• If the reference angle in Question 1 is B, Find the adjacent ?
• If the reference angle in Question 2 is Z, find the opposite?

Use your calculator to find the given ratios. Round to four decimal places if necessary:

For the given triangles, find the sine, cosine and tangent of the θ.

For the given triangles, find the missing side. Round it to one decimal place.

For the given triangles, find the missing sides. Round it to one decimal place.

Solve the triangles. Round to one decimal place.

• Kim stands 75 metres from the bottom of a tree and looks up at the top of the tree at a 48° angle. How tall is the tree?
• A tree makes a shadow that is 6 metres long when the angle of elevation to the sun is 52°. How tall is the tree?
• A ladder that is 15 feet is leaning against a house and makes a 45° angle with the ground. How far is the base of the ladder from the house?
• Roxanne is flying a kite and has let out 100 feet of string. The angle of elevation with the ground is 38°. How high is her kite above the ground?
• Marta is flying a kite and has let out 28 metres of string. If the kite is 10 metres above the ground, what is the angle of elevation?
• An airplane takes off from the ground at the angle of 25°. If the airplane traveled 200 kilometres, how high above the ground is it?

• y = 19.3, z = 8.2

Attribution:

This chapter has been adapted from “Solve Applications: Sine, Cosine and Tangent Ratios” in Introductory Algebra  by Izabela Mazur, which is under a CC BY 4.0 Licence . See the Adaptation Statement for more information.

Business/Technical Mathematics Copyright © 2021 by Izabela Mazur and Kim Moshenko is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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• Preparatory

Lesson Explainer: Evaluating Trigonometric Functions with Special Angles Mathematics • First Year of Secondary School

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• Remaining Seats: 8

In this explainer, we will learn how to evaluate the trigonometric functions with special angles and how to use them to evaluate trigonometric expressions.

We will begin by recalling special angles, together with the sine, cosine, and tangent values of these angles.

Let us consider a unit circle. This enables us to calculate s i n 𝑥 , c o s 𝑥 , and t a n 𝑥 between 0 ∘ and 3 6 0 ∘ or 0 and 2 𝜋 radians . All three functions have key values at 0 ∘ , 9 0 ∘ , 1 8 0 ∘ , 2 7 0 ∘ , and 3 6 0 ∘ . We know that 1 8 0 = 𝜋 ∘ r a d i a n s , so 3 6 0 = 2 𝜋 ∘ r a d i a n s , 9 0 = 𝜋 2 ∘ r a d i a n s , and 2 7 0 = 3 𝜋 2 ∘ r a d i a n s . Knowing these conversions enables us to solve trigonometric problems when the angles are given in either degrees or radians .

As the functions are periodic, we can calculate the sine, cosine, and tangent of angles outside of this range by adding multiples of 3 6 0 ∘ for s i n and c o s , or by adding multiples of 1 8 0 ∘ for t a n .

Next, we recall that the special angles are 3 0 ∘ , 4 5 ∘ , and 6 0 ∘ . The sine, cosine, and tangent of these angles are given below.

While we will not consider the derivation or proof of these results in this explainer, it is worth recalling the identity s i n c o s t a n 𝑥 𝑥 = 𝑥 . This enables us to calculate the tangent of any angle if we are given the sine and cosine of that angle.

For example, since s i n 6 0 = √ 3 2 ∘ and c o s 6 0 = 1 2 ∘ , then t a n 6 0 = √ 3 2 ÷ 1 2 = √ 3 ∘ .

It will also be necessary for us to recall the definitions of the reciprocal trigonometric functions.

Definition: Reciprocal Trigonometric Functions

The reciprocal trigonometric functions cosecant 𝜃 , secant 𝜃 , and cotangent 𝜃 are the reciprocal of sine 𝜃 , cosine 𝜃 , and tangent 𝜃 such that c s c s i n s e c c o s c o t t a n 𝜃 = 1 𝜃 , 𝜃 = 1 𝜃 , 𝜃 = 1 𝜃 .

We can use these identities to calculate the cosecant, secant, and cotangent of 3 0 ∘ , 4 5 ∘ , and 6 0 ∘ .

We will now recall the related angles of the trigonometric functions:

One way of recalling whether the sine, cosine, and tangent of any angle between 0 ∘ and 3 6 0 ∘ are positive or negative is using the CAST diagram. This is a memory device that we use to remember the signs of the trigonometric ratios in each of the four quadrants.

We note that the angles are measured, from 0 ∘ to 3 6 0 ∘ or from 0 to 2 𝜋 radians in a counterclockwise direction, where the positive 𝑥 -axis is the initial side of the angle. The terminal side is where the angle stops. Any angle between 0 ∘ and 9 0 ∘ lies in the first quadrant. Any angle between 9 0 ∘ and 1 8 0 ∘ lies in the second quadrant. Any angle between 1 8 0 ∘ and 2 7 0 ∘ lies in the third quadrant. Any angle between 2 7 0 ∘ and 3 6 0 ∘ lies in the fourth quadrant.

Let us consider an example where we have to evaluate the cosine of an angle using the property of related angles to relate it to a special angle.

Example 1: Using Periodic Identities to Find the Value of a Trigonometric Function Involving Special Angles

Find the value of c o s 1 1 𝜋 6 .

We begin by recalling that 𝜋 = 1 8 0 r a d i a n s ∘ .

So, 𝜋 6 = 3 0 1 1 𝜋 6 = 3 3 0 . r a d i a n s r a d i a n s ∘ ∘

We therefore need to calculate c o s 3 3 0 ∘ .

Let us recall the property of related angles c o s c o s ( 3 6 0 − 𝜃 ) = 𝜃 . ∘

If 𝜃 = 3 0 ∘ , then c o s c o s c o s c o s ( 3 6 0 − 3 0 ) = 3 0 3 3 0 = 3 0 . ∘ ∘ ∘ ∘ ∘

From our knowledge of special angles, we know that c o s 3 0 = √ 3 2 ∘ .

So, c o s 3 3 0 = √ 3 2 . ∘

We can therefore conclude that c o s 1 1 𝜋 6 = √ 3 2 .

Let us now consider a similar example where, this time, we need to evaluate the sine and tangent of given angles.

Example 2: Evaluating Trigonometric Expressions Involving Special Angles

Evaluate 2  𝜋 6  − 8  4 𝜋 3  t a n s i n .

So, 𝜋 6 = 3 0 . r a d i a n s ∘

Also, 𝜋 3 = 6 0 4 𝜋 3 = 2 4 0 . r a d i a n s r a d i a n s ∘ ∘

We therefore need to calculate 2 3 0 − 8 2 4 0 t a n s i n ∘ ∘ .

From our knowledge of special angles, we know that t a n 3 0 = √ 3 3 ∘ and s i n 6 0 = √ 3 2 ∘ .

By considering the CAST diagram, as shown below, we see that 2 4 0 ∘ lies in the 3rd quadrant and the sine of any angle here is negative.

Since s i n s i n ( 1 8 0 + 𝜃 ) = − 𝜃 , ∘ then s i n s i n s i n s i n ( 1 8 0 + 6 0 ) = − 6 0 ( 2 4 0 ) = − 6 0 . ∘ ∘ ∘ ∘ ∘

So, s i n ( 2 4 0 ) = − √ 3 2 . ∘

Substituting the values of t a n 3 0 ∘ and s i n 2 4 0 ∘ into our expression, we have 2 3 0 − 8 2 4 0 = 2  √ 3 3  − 8  − √ 3 2  = 2 √ 3 3 + 4 √ 3 = 2 √ 3 3 + 1 2 √ 3 3 = 1 4 √ 3 3 . t a n s i n ∘ ∘

Therefore, 2 𝜋 6 − 8 4 𝜋 3 = 1 4 √ 3 3 t a n s i n .

In the remaining examples in this explainer, we will also need to use reciprocal trigonometric functions.

Example 3: Using Periodic Identities to Evaluate a Trigonometric Expression Involving Special Angles

Find the value of c o s t a n c s c c o s 1 3 5 + 1 3 5 + 2 2 5 + 2 2 5 ∘ ∘ ∘ ∘ .

In order to answer this question, we need to recall reciprocal trigonometric identities and special angles.

From our knowledge of special angles, s i n 4 5 = √ 2 2 ∘ , c o s 4 5 = √ 2 2 ∘ , and t a n 4 5 = 1 ∘ .

Using the CAST diagram, we know that both the cosine and tangent of any angle in the second quadrant are negative. The sine and cosine of any angle in the third quadrant are also both negative.

The properties of related angles state that c o s c o s ( 1 8 0 − 𝜃 ) = − 𝜃 . ∘

So, c o s c o s c o s c o s ( 1 8 0 − 4 5 ) = − 4 5 1 3 5 = − 4 5 = − √ 2 2 . ∘ ∘ ∘ ∘ ∘

The properties of related angles also state that t a n t a n ( 1 8 0 − 𝜃 ) = − 𝜃 . ∘

So, t a n t a n ( 1 8 0 − 4 5 ) = − 4 5 . ∘ ∘ ∘

Therefore, t a n t a n 1 3 5 = − 4 5 = − 1 . ∘ ∘

They also state that s i n s i n ( 1 8 0 + 𝜃 ) = − 𝜃 . ∘

So, s i n s i n ( 1 8 0 + 4 5 ) = − 4 5 . ∘ ∘ ∘

Therefore, s i n s i n 2 2 5 = − 4 5 = − √ 2 2 . ∘ ∘

They also state that c o s c o s ( 1 8 0 + 𝜃 ) = − 𝜃 . ∘

So, c o s c o s ( 1 8 0 + 4 5 ) = − 4 5 . ∘ ∘ ∘

Therefore, c o s c o s 2 2 5 = − 4 5 = − √ 2 2 . ∘ ∘

Since c s c s i n 𝜃 = 1 𝜃 , then c s c s i n 2 2 5 = 1 2 2 5 = − √ 2 . ∘ ∘

Substituting these values into our expression, we have c o s t a n c s c c o s 1 3 5 + 1 3 5 + 2 2 5 + 2 2 5 = − √ 2 2 + ( − 1 ) +  − √ 2  +  − √ 2 2  = − √ 2 2 − 1 − √ 2 − √ 2 2 = − 2 √ 2 − 1 . ∘ ∘ ∘ ∘

So, our final answer is − 2 √ 2 − 1 .

For the next example, we will need to evaluate a second-degree trigonometric expression.

Example 4: Evaluating Trigonometric Expressions Involving Special Angles

Evaluate 3 − 4 3 3 0 c o s  ∘ .

Substituting this value into our expression gives us 3 − 4 3 3 0 = 3 − 4  √ 3 2  = 3 − 4  3 4  = 3 − 3 = 0 . c o s  ∘ 

So, our final answer is 0.

In the next example, we will investigate how different angles can satisfy a trigonometric equation involving multiple angles and the power of a trigonometric function.

Example 5: Using Periodic Identities to Evaluate a Trigonometric Expression Involving Special Angles

Which of the following values of 𝑥 does not satisfy the equation s i n c o s  𝑥 + 6 ( 3 𝑥 ) − 1 2 = − 3 √ 2 ?

In this example, we have been given several values of 𝑥 and are being asked which does not satisfy the given equation. We note that since the given equation is a trigonometric one, it can possibly have an infinite number of solutions due to the properties of related angles.

The easiest way to check which value of 𝑥 does not satisfy the equation is to substitute the given values into the left-hand side one by one and see whether we get the right-hand side.

First, for 𝑥 = 4 5 ∘ , we have s i n c o s  ∘ ∘ 4 5 + 6 ( 3 ⋅ 4 5 ) − 1 2 .

To evaluate this, we first recall that 4 5 ∘ is a special angle, and so we can look up the value of the first term directly to find that s i n 4 5 = √ 2 2 ∘ . For the second term, that is, 6 ( 3 ⋅ 4 5 ) = 6 1 3 5 c o s c o s ∘ ∘ , we can use the related angle property: c o s c o s ( 1 8 0 − 𝜃 ) = − 𝜃 . ∘

Taking 𝜃 = 4 5 ∘ , we have c o s c o s 1 3 5 = − 4 5 = − √ 2 2 . ∘ ∘

Putting this together, we have s i n c o s  ∘ ∘  4 5 + 6 ( 1 3 5 ) − 1 2 =  √ 2 2  + 6  − √ 2 2  − 1 2 = 2 4 − 6 √ 2 2 − 1 2 = 1 2 − 3 √ 2 − 1 2 = − 3 √ 2 .

Thus, since this is equal to the right-hand side, 4 5 ∘ satisfies the equation.

Let us repeat this process for the other angles, continuing to use properties of related angles to help us. For 𝑥 = 1 3 5 ∘ , we have s i n c o s  ∘ ∘ 1 3 5 + 6 4 0 5 − 1 2 .

We use the identities s i n s i n ( 1 8 0 − 𝜃 ) = 𝜃 ∘ and c o s c o s ( 𝜃 + 3 6 0 ) = 𝜃 ∘ , with 𝜃 = 4 5 ∘ , to get s i n c o s s i n c o s  ∘ ∘  ∘ ∘  1 3 5 + 6 4 0 5 − 1 2 = 4 5 + 6 4 5 − 1 2 =  √ 2 2  + 6  √ 2 2  − 1 2 = 1 2 + 6 √ 2 2 − 1 2 = 3 √ 2 .

As this is not equal to the right-hand side, this shows that 1 3 5 ∘ does not satisfy the equation.

We have already found the answer, but for completeness, let us check the remaining options. For 𝑥 = 3 1 5 ∘ , we have s i n c o s  ∘ ∘ 3 1 5 + 6 9 4 5 − 1 2 .

The first term can be simplified using s i n s i n ( 3 6 0 − 𝜃 ) = − 𝜃 ∘ (by taking 𝜃 = 4 5 ∘ ), to get s i n s i n 3 1 5 = − 4 5 ∘ ∘ . The second can be simplified by first using the periodicity of cosine, in other words, c o s c o s ( 𝜃 + 2 ⋅ 3 6 0 ) = 𝜃 ∘ , to get c o s c o s 9 4 5 = 2 2 5 ∘ ∘ , and then using c o s c o s ( 3 6 0 − 𝜃 ) = 𝜃 ∘ to get c o s c o s 2 2 5 = 1 3 5 ∘ ∘ . Recall that we have already found that c o s 1 3 5 = − √ 2 2 ∘ . Putting this together, we have s i n c o s s i n c o s  ∘ ∘ ∘  ∘  ( 3 1 5 ) + 6 ( 9 4 5 ) − 1 2 = ( − 4 5 ) + 6 1 3 5 − 1 2 =  − √ 2 2  + 6  − √ 2 2  − 1 2 = 1 2 − 3 √ 2 − 1 2 = − 3 √ 2 .

Thus, 𝑥 = 3 1 5 ∘ works. The fourth option is 𝑥 = 4 0 5 ∘ . For this, we have s i n c o s  ∘ ∘ 4 0 5 + 6 1 2 1 5 − 1 2 .

For both the first and second terms, we can use the periodicity of sine and cosine to simplify these terms. That is, s i n s i n ( 𝜃 + 3 6 0 ) = 𝜃 ∘ and c o s c o s ( 𝜃 + 3 ⋅ 3 6 0 ) = 𝜃 ∘ to get s i n c o s s i n c o s  ∘ ∘  ∘ ∘ 4 0 5 + 6 1 2 1 5 − 1 2 = 4 5 + 6 1 3 5 − 1 2 .

This is the same left-hand side as we had for 𝑥 = 4 5 ∘ , so we can conclude that this value satisfies the equation.

Finally, we have 𝑥 = − 3 1 5 ∘ , which gives us a left-hand side of s i n c o s  ∘ ∘ ( − 3 1 5 ) + 6 ( − 9 4 5 ) − 1 2 .

We can use the periodic identities s i n s i n 𝜃 = ( 𝜃 + 3 6 0 ) ∘ and c o s c o s 𝜃 = ( 𝜃 + 3 ⋅ 3 6 0 ) ∘ to get s i n c o s s i n c o s  ∘ ∘  ∘ ∘ ( − 3 1 5 ) + 6 ( − 9 4 5 ) − 1 2 = 4 5 + 6 1 3 5 − 1 2 .

Once more, we have the same left-hand side as we had for 𝑥 = 4 5 ∘ and 𝑥 = 4 0 5 ∘ , so we can conclude that this value satisfies the equation.

In conclusion, option B does not satisfy the equation.

In the previous example, let us note that three of the possible solutions, namely, 𝑥 = 4 5 ∘ , 𝑥 = 4 0 5 ∘ , and 𝑥 = − 3 1 5 ∘ , ended up with the exact same expression on the left-hand side. This was a result of the periodicity of trigonometric functions, which can be seen by considering the positions of these angles on a CAST diagram.

That is to say, the values of c o s 𝑥 and s i n 𝑥 are equal for these values of 𝑥 because they share the same place on the CAST diagram. By extension, the values of c o s  𝑥 and s i n  𝑥 will also be the same.

It is important to be aware, however, that we cannot always use this approach of a CAST diagram. In particular, we should be careful since one of the terms in the equation (i.e., 6 ( 3 𝑥 ) c o s ) has a multiple angle. Recall that for trigonometric functions with multiple angles, the periodicity is different. This difference is illustrated in the graphs below.

Specifically, the period of c o s 𝑥 is 3 6 0 ∘ (or 2 𝜋 ), but the period of c o s 3 𝑥 is 1 2 0 ∘ (or 2 𝜋 3 ). Nevertheless, since the periods of c o s 𝑥 and c o s 3 𝑥 overlap, it turns out that angles that are 3 6 0 ∘ apart will still have the same value of c o s 3 𝑥 . We can see this by considering the periodic property of c o s 3 𝑥 , specifically that since it has a period of 1 2 0 ∘ , we have c o s c o s 3 𝑥 = ( 3 ( 𝑥 + 1 2 0 𝑛 ) ) , ∘ where 𝑛 is an integer. Since 3 6 0 ∘ is three times 1 2 0 ∘ , if we let 𝑛 = 3 𝑘 , where 𝑘 is also an integer, we have c o s c o s c o s 3 𝑥 = ( 3 ( 𝑥 + 1 2 0 ⋅ 3 𝑘 ) ) = ( 3 ( 𝑥 + 3 6 0 𝑘 ) ) . ∘ ∘

This shows us that the value of c o s 3 𝑥 will be the same for any angles that differ by 3 6 0 ∘ , so 4 5 ∘ , 4 0 5 ∘ , and − 3 1 5 ∘ will be the same too.

For our final example, let us find the value of an expression involving the product of multiple trigonometric functions.

Example 6: Evaluating Trigonometric Expressions Involving Special Angles

Find the value of 3 3 0 6 0 − 0 6 0 + 2 7 0 4 5 s i n s i n c o s s e c s i n c o s ∘ ∘ ∘ ∘ ∘  ∘ .

From our knowledge of special angles, we know that s i n 3 0 = 1 2 ∘ , s i n 6 0 = √ 3 2 ∘ , and c o s 4 5 = √ 2 2 ∘ .

From the sine and cosine graphs below, we see that c o s 0 = 1 ∘ and s i n 2 7 0 = − 1 ∘ .

Since s e c c o s 𝜃 = 1 𝜃 , then s e c c o s 6 0 = 1 6 0 = 1 ÷ 1 2 = 2 . ∘ ∘

We can now substitute all of these values into our expression: 3 3 0 6 0 − 0 6 0 + 2 7 0 4 5 = 3  1 2   √ 3 2  − ( 1 ) ( 2 ) + ( − 1 )  √ 2 2  = 3 √ 3 4 − 2 − 1 2 = 3 √ 3 4 − 1 0 4 = − 1 0 + 3 √ 3 4 . s i n s i n c o s s e c s i n c o s ∘ ∘ ∘ ∘ ∘  ∘ 

We will finish this explainer by recapping some of the key points.

• We can evaluate trigonometric functions and expressions using our knowledge of special angles: s i n s i n s i n c o s c o s c o s t a n t a n t a n 3 0 = 1 2 , 4 5 = √ 2 2 , 6 0 = √ 3 2 , 3 0 = √ 3 2 , 4 5 = √ 2 2 , 6 0 = 1 2 , 3 0 = 1 √ 3 , 4 5 = 1 , 6 0 = √ 3 . ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘
• We can use reciprocal trigonometric identities to solve more complicated problems: c s c s i n s e c c o s c o t t a n 𝜃 = 1 𝜃 , 𝜃 = 1 𝜃 , 𝜃 = 1 𝜃 .

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