steps in problem solving involving trigonometric ratios brainly

• Secondary School

Show complete solution Steps in problem-solving involving trigonometric ratios: 1. Draw a diagram. Transform it into a geometric triangle and label each part correctly. 2. Determine the given and the question that needs to be answered. 3. Identify the corresponding trigonometric formula to be used. 4. Show your solution. 5. Present the conclusion.

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Trig functions

Here you will learn what the trigonometric functions and inverse trigonometric functions are and how you can use them to calculate missing side lengths and angles in right triangles.

Students will first learn about trig functions as part of geometry in high school.

What are trig functions?

Trig functions , or trigonometric functions, are functions that relate an angle in a right triangle to the ratio of two of its sides.

Sin, Cos, and Tan (sine, cosine, tangent) are trig functions.

The angle is often referred to as theta and is represented using the symbol \theta.

sin θ (The sine of angle theta)

When you input a given angle into \sin\theta, it gives you the ratio between the opposite length and the hypotenuse.

It tells you how many times greater the opposite side is than the hypotenuse.

[FREE] Trigonometry Worksheet (Grade 9 to 12)

Use this quiz to check your grade 9 to 12 students’ understanding of Trigonometry. 15+ questions with answers covering a range of 9th to 12th grade trigonometry topics to identify areas of strength and support!

cos θ (The cosine of angle theta)

When you input a given angle into \cos\theta , it gives you the ratio between the adjacent length and the hypotenuse.

It tells you how many times greater the adjacent side is than the hypotenuse.

tan θ (The tangent of angle theta)

When you input a given angle into \tan\theta , it gives you the ratio between the adjacent length and the opposite length.

It tells you how many times greater the opposite is than the adjacent.

In summary, trigonometric functions can tell you how many times greater one side is than another in triangles containing a right angle. Therefore, if you are given the length of one side and the value of one angle, you can set up equations using trigonometric functions to help you find the lengths of the other sides.

Alternatively, if you are given the length of two sides of a right triangle, you can determine the size of the respective angle. This is calculated using inverse trigonometric functions.

What are inverse trigonometric functions?

In the same way that addition and subtraction are inverse operations, inverse trig functions do the opposite of regular trig functions.

You can use the inverse sine function , the inverse cosine function , and the inverse tangent function to find the missing angle \theta.

The inverse trig functions are:

The regular trig functions are functions that act on an angle \theta.

As these functions are equal to the ratio of two respective side lengths, the inverse trig functions act on this ratio, not the angle.

It is important to note that “trig functions” is an overarching term for all trigonometric functions, including inverse trigonometric functions. This means that “inverse trig functions” can be used interchangeably with “trig functions”.

Common Core State Standards

How does this relate to high school math?

• High School – Geometry – Similarity, Right Triangles, & Trigonometry (HS.G.SRT.C.6) Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.
• High School – Geometry – Similarity, Right Triangles, & Trigonometry (HS.G.SRT.C.8) Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

How to use trigonometric functions

In order to use trigonometric functions:

Label one angle in the right triangle that is known/required. This cannot be the right-angle.

Label the side lengths with respect to the known/required angle and reject one side that is unknown/not useful.

Set up an equation involving the correct trig function.

Rearrange the equation so the subject is the value you are going to find.

Find the answer using a calculator and round it as needed.

Write the answer with the correct units.

Trig functions examples

Example 1: calculating with trigonometric functions.

Find the value of x, given \sin(72)=\cfrac{25}{x}. Give your answer to the nearest tenth.

You already have an equation involving \sin , so there is no angle to label.

2 Label the side lengths with respect to the known/required angle and reject one side that is unknown/not useful.

3 Set up an equation involving the correct trig function.

The given equation is \sin(72)=\cfrac{25}{x} .

4 Rearrange the equation so the subject is the value you are going to find.

Multiply both sides of the equation by x first.

Divide both sides of the equation by \sin(72) .

5 Find the answer using a calculator and round it as needed.

Type \cfrac{25}{\sin(72)} into a calculator to find x.

x=\cfrac{25}{\sin(72)}=26.28655561=26.2 (nearest tenth)

6 Write the answer with the correct units.

There are no units stated, however it is important to note that the value of x would represent a length and so units may be \mathrm{cm, km, in} etc, not an angle with the units ^{\circ}.

Example 2: finding the length of a side

Find the length labeled x in the triangle below. Give your answer to 3 significant figures.

The known angle that is not the right angle is 34^{\circ}.

Labeling the triangle, you can see that you know the hypotenuse H and you want to know the adjacent, A. Reject the opposite side labeled O. The trigonometric function which compares these two sides is \cos.

You know that \cos(\theta)=\cfrac{\text{Adjacent}}{\text{Hypotenuse}} , therefore \cos(34)=\cfrac{x}{32}.

Multiply both sides of the equation by 32 .

32\times\cos(34)=x

Example 3: finding the length of a side

The known angle that is not the right angle is 29^{\circ}.

Labeling the triangle, you can see that you know the opposite and you want to know the adjacent. The trigonometric function which compares these two sides is \tan.

You know that \tan(\theta)=\cfrac{\text{Opposite}}{\text{Adjacent}} , therefore \tan(29)=\cfrac{8}{x}.

Multiply both sides of the equation by x .

x\times\tan(29)=8

Divide both sides of the equation by \tan(29) .

x=\cfrac{8}{\tan(29)}

Example 4: use inverse trigonometric functions

Find the value of \theta to the nearest tenth.

You already have an equation involving \sin so there is no angle to label.

\sin(\theta)=0.9 is already in the required form.

Here, the question requires you to find the angle \theta. As the sine function is currently acting on \theta , use the inverse sine function on both sides of the equation to make \theta the subject.

\begin{aligned}\sin^{-1}(\sin(\theta))&=\sin^{-1}(0.9) \\\\ \theta&=\sin^{-1}(0.9) \end{aligned}

Note: See the practical analogy at the end of the example for help with understanding why \sin^{-1}(\sin(\theta))=\theta

\theta=64.15806724=64.2 (nearest tenth)

A useful analogy to help understand \sin^{-1}(\sin(\theta))=\theta\text{:}

Imagine throwing a ball straight up in the air, and then catching it in exactly the same place the ball started.

Here, the ball is \theta.

Let the function \sin represent throwing the ball (\theta).

Let the function \sin^{-1} represent catching the ball (\theta).

Throwing the ball is an action on the ball. Catching the ball is the opposite action on the ball. If you throw the ball, then catch the ball in exactly the same place, nothing about the ball or its position have changed as both of the acts on the ball have canceled each other out. In essence, nothing has happened.

So catch(throw(ball)) = ball or \sin^{-1}(\sin(\theta))=\theta

Example 5: use inverse trigonometric functions

Find the value of \theta to the nearest hundredth when 2\cos(\theta)=0.8.

You already have an equation involving \cos , so there is no angle to label.

You already have an equation involving \cos , so there are no side lengths to label.

You have the equation 2\cos(\theta)=0.8

You want to isolate \theta on one side of the equation, so first divide both sides of the equation by 2.

\cos(\theta)=0.4

Apply the inverse cosine trigonometric function to each side of the equation to make \theta the subject.

\theta=\cos^{-1}(0.4)

\theta=66.42182152=66.42 (nearest hundredth)

\theta=66.42^{\circ} (nearest hundredth)

Example 6: calculate an angle in a triangle

Calculate the unknown angle x to 1 significant figure.

The required angle is x^{\circ}.

Labeling the triangle, you can see that you know the opposite and you know the adjacent. The trigonometric function which compares these two sides is \tan.

You know that \tan(x)=\cfrac{\text{Opposite}}{\text{Adjacent }}

So, \tan(x)=\cfrac{8}{14.2}

Apply the inverse tangent trigonometric function to each side of the equation to make x the subject.

x=\tan^{-1}\left(\cfrac{8}{14.2}\right)

You can check that the answer is sensible here. All angles (other than the right angle) in a right triangle should be acute angles. 30^{\circ} is acute therefore the answer is sensible.

Teaching tips for trig functions

• Introduce the unit circle to explain how the values of sine and cosine change as the angle increases. Stress that the radius of the unit circle is 1, so sine and cosine are just the y – and x -coordinates of a point on the circle.
• Once students are familiar with the basic trigonometric functions, connect them to the Pythagorean identities. Show the derivation of \sin^{2}(\theta)+\cos^{2}(\theta)=1 from the Pythagorean theorem, reinforcing the geometric origins of trigonometry.
• Emphasize symmetry in trigonometric functions, such as how sine is an odd function and cosine is even. Show students the periodic nature of these functions and how they repeat every 360^{\circ}.

Easy mistakes to make

• Using the wrong function You use sin, cos, and tan to find side lengths and inverse sin, inverse cos, and inverse tan to find angles.
• Not knowing other trig identities There are other trigonometric identities as well as sin, cos, and tan, called reciprocal identities. These include the cosecant function, the secant function and the cotangent function. \cfrac{1}{\sin\theta}=\mathrm{cosecant} \, \theta\text{ (or }\csc\theta) \cfrac{1}{\cos\theta}=\mathrm{secant} \, \theta\text{ (or }\sec\theta) \cfrac{1}{\tan\theta}=\mathrm{cotangent} \, \theta\text{ (or }\cot\theta)
• \sin^{-1}(x) as \arcsin(x)
• \cos^{-1}(x) as \arccos(x)
• \tan^{-1}(x) as \arctan(x) avoids this confusion!
• Using degrees instead of radians (or vice versa) Encourage students to always check whether they are working in degrees or radians. Using a visual unit circle for both units can help solidify the connection.
• Forgetting to change signs depending on the quadrant Students may forget that trigonometric functions change signs depending on the quadrant (for example, sine is positive in the 1 st and 2 nd quadrants but negative in the 3 rd and 4 th).

Related trigonometry lessons

• Trigonometry
• How to find the exact value of a trig function
• Sin cos tan
• Law of sines
• Trig formula for area of a triangle
• Trig formulas
• Trig identities
• Law of cosines

Practice trig functions questions

1. Find the value of x when \tan{42}=\cfrac{x}{5}, giving your answer to the nearest tenth.

2. Find the length of the side labeled a on this triangle, giving your answer to 2 significant figures.

Label the known angle that is not the right angle:

Label the side lengths, reject the irrelevant side:

The trigonometric function involving the opposite side and the hypotenuse is \sin\theta=\cfrac{O}{H}.

As O=16\mathrm{~cm}, \, H=a and \theta=61^{\circ}

Rearranging this you get

3. Find the length AB. Give your answer to the nearest hundredth.

The trigonometric function involving the opposite side and the adjacent side is \tan\theta=\cfrac{O}{A}.

As O=AB, \, A=4\mathrm{~cm}   and \theta=73^{\circ}

4. Find the value of x when \cos(x)=0.2. Give your answer to the nearest tenth.

\cos(x)=0.2 so x=\cos^{-1}(0.2)=78.5^{\circ} (nearest tenth)

5. Solve the equation \sin(\theta)+5=5.7. Give your answer to the nearest tenth.

6. Find the value of \theta in the triangle below. Give your answer to the nearest tenth.

First you need to label the required sides and angles, and set up the equation using SOHCAHTOA.

You know the opposite and the hypotenuse so you need to use \sin.

Trig functions FAQs

The three basic trigonometric functions are sine (\sin), cosine (\cos), and tangent (\tan).

The reciprocal functions are cosecant (\csc), secant (\sec), and cotangent (\cot).

The unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. Trig functions for an angle \theta can be defined using the coordinates of points on the unit circle: ° \sin (\theta) is the y -coordinate ° \cos (\theta) is the x -coordinate The unit circle also helps determine values of trig functions for any angle, not just those in a right triangle.

Periodic functions repeat their values at regular intervals, called the period. Trigonometric functions like sine and cosine are periodic because their values repeat every 360^{\circ}, while tangent repeats every 180^{\circ}. This periodicity is key to understanding the cyclical nature of trig functions.

The next lessons are

• Circle math
• Sectors arcs and segments
• Circle theorems

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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  

• 10th Grade Math

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Solve Problems Using Trigonometric Ratios

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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Chapter 2: Trigonometric Ratios

2.3 Solving Right Triangles

Algebra refresher.

1. [latex]\:\sqrt{2}\sqrt{2}[/latex]

2. [latex]\:\dfrac{3}{\sqrt{3}}[/latex]

3. [latex]\:\sqrt{8}[/latex]

4. [latex]\:\sqrt{\dfrac{3}{4}}[/latex]

Rationalize the denominator.

5. [latex]\:\dfrac{1}{\sqrt{2}}[/latex]

6. [latex]\:\dfrac{2}{\sqrt{3}}[/latex]

7. [latex]\:\dfrac{6}{\sqrt{3}}[/latex]

8. [latex]\:\dfrac{4}{\sqrt{8}}[/latex]

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

• [latex]\displaystyle 2\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]
• [latex]\displaystyle \sqrt{3}\vphantom{\vphantom{\dfrac{\sqrt{3}}{2}}}[/latex]
• [latex]\displaystyle 2\sqrt{2}\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]
• [latex]\displaystyle \dfrac{\sqrt{3}}{2}[/latex]
• [latex]\displaystyle \dfrac{\sqrt{2}}{2}[/latex]
• [latex]\displaystyle \dfrac{2\sqrt{3}}{3}[/latex]
• [latex]\displaystyle 2\sqrt{3}\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]
• [latex]\displaystyle \sqrt{2}\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]

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 [latex]90 {^o}\text{.}[/latex] 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 [latex]35 {^o}\text{,}[/latex] as shown in the figure. Solve the triangle.

We can find the side opposite the 35° angle by using the sine ratio. [latex]\begin{align*} \sin 35 {^o} = \dfrac {\text{opposite}}{\text{hypotenuse}}\\ 0.5736 = \dfrac{a}{150}\\ a = 150(0.5736) = 86.04 \end{align*}[/latex] The opposite side is about 86 feet long. To find side [latex]b[/latex], 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. [latex]\begin{align*} \cos 35 {^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}}\\ 0.8192 = \dfrac{b}{150}\\ b = 150(0.8192) = 122.89 \end{align*}[/latex] The adjacent side is about [latex]123[/latex] feet long. Finally, the unknown angle is the complement of [latex]35 {^o}\text{,}[/latex] or [latex]55 {^o}\text{.}[/latex]

Checkpoint 2.32.

Sketch a right triangle with

• one angle of [latex]37 {^o}\text{,}[/latex] 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 [latex]37{^o}[/latex] to find the opposite side. Use cos [latex]37{^o}[/latex] to find the hypotenuse. Subtract [latex]37{^o}[/latex] from [latex]90{^o}[/latex] 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 [latex]\begin{equation*} \dfrac {\Delta y}{\Delta x} = \dfrac{77}{136} = 0.5662 \end{equation*}[/latex] 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 [latex]\boxed{\text{TAN}^{-1}}[/latex] on your calculator; it is probably the second function above the TAN key. Enter [latex]\qquad\qquad\qquad[/latex] 2nd TAN 0.5662

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

When we find [latex]\tan^{-1}[/latex] of a number, we are finding an angle whose tangent is that number. Similarly, [latex]\sin^{-1}[/latex] and [latex]cos^{-1}[/latex] 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 [latex]0.6834\text{.}[/latex]

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

In the last example, the two equations:

[latex]\begin{equation*} \sin 43.11{^o} = 0.6834\end{equation*}[/latex]

[latex]\begin{equation*} \sin^{-1} 0.6834 = 43.11{^o}\end{equation*}[/latex]

say the same thing in different ways.

Caution 2.35.

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

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

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

Checkpoint 2.36.

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

[latex]\cos 68{^o} = 0.3746[/latex] or [latex]\cos^{-1}(0.3746) = 68{^o}[/latex]

Example 2.37.

Find the angle of inclination of a hill if you gain [latex]400[/latex] feet in elevation while traveling half a mile.

A sketch of the hill is shown at right. (Recall that [latex]1[/latex] mile [latex]= 5280[/latex] feet.)

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

The angle of inclination of the hill is about [latex]8.7{^o}\text{.}[/latex]

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.

[latex]75{^o}[/latex]

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 [latex]30{^o}\text{.}[/latex] 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 [latex]30{^o}[/latex] angle is half the length of the hypotenuse, so [latex]\sin 30{^o} = \dfrac{1}{2}\text{.}[/latex]

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

Now we know the cosine and tangent of [latex]30{^o}\text{.}[/latex] [latex]\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*}[/latex]

These are exact values for the trig ratios, but we can also find decimal approximations. Use your calculator to verify the following approximate values. [latex]\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*}[/latex]

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 [latex]30{^o}\text{,}[/latex] [latex]60{^o}\text{,}[/latex] and [latex]45{^o}[/latex] 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 [latex]8[/latex] centimeters long. Find the exact length of the triangle’s altitude, [latex]h\text{.}[/latex]

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

[latex]\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*}[/latex] The altitude is exactly [latex]4\sqrt{3}[/latex] centimeters long.

From this exact answer, we can find approximations to any degree of accuracy we like. You can check that [latex]4\sqrt{3} \approx6.9282\text{,}[/latex] 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 [latex]60{^o}\text{.}[/latex]

[latex]\sin 60{^o} = \dfrac{\sqrt{3}}{2}\text{,}[/latex] [latex]\cos 60{^o} = \dfrac{1}{2}\text{,}[/latex] [latex]\tan 60{^o} = \sqrt{3}[/latex]

There is one more special angle: [latex]45{^o}\text{.}[/latex] 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 [latex]45{^o}\text{.}[/latex] 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 [latex]\sqrt{2}\text{,}[/latex] so the trig ratios for [latex]45{^o}[/latex] are[latex]\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*}[/latex]

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 [latex]60{^o} \approx 0.8660\text{,}[/latex] so if sin [latex]\theta = 0.95[/latex] for some unknown angle [latex]\theta\text{,}[/latex] we know that [latex]\theta \gt 60{^o}\text{,}[/latex] because as [latex]\theta[/latex] increases from [latex]60{^o}[/latex] to [latex]90{^o}[/latex] , the sine of [latex]\theta[/latex] increases also.

Example 2.43.

If [latex]\cos \alpha \gt \dfrac{\sqrt{3}}{2}\text{,}[/latex] what can we say about [latex]\alpha\text{?}[/latex]

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

Checkpoint 2.44.

If [latex]1 \lt \tan \beta \lt \sqrt{3}\text{,}[/latex] what can we say about [latex]\beta\text{?}[/latex]

[latex]45{^o} \lt \beta \lt 60{^o}[/latex]

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 [latex]\sin^{-1}[/latex] (or [latex]\cos^{-1}[/latex] or [latex]\tan^{-1}[/latex]) 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.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Study Guides > Precalculus II

Solving trigonometric equations, solving linear trigonometric equations in sine and cosine, example 1: solving a linear trigonometric equation involving the cosine function, example 2: solving a linear equation involving the sine function, how to: given a trigonometric equation, solve using algebra..

• Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
• Substitute the trigonometric expression with a single variable, such as [latex]x[/latex] or [latex]u[/latex].
• Solve the equation the same way an algebraic equation would be solved.
• Substitute the trigonometric expression back in for the variable in the resulting expressions.
• Solve for the angle.

Example 3: Solve the Trigonometric Equation in Linear Form

Solve trigonometric equations using a calculator, example 8: using a calculator to solve a trigonometric equation involving sine, analysis of the solution, example 9: using a calculator to solve a trigonometric equation involving secant.

Solving Equations Involving a Single Trigonometric Function

Example 4: solving a problem involving a single trigonometric function, example 5: solving a trigonometric equation involving cosecant, example 6: solving an equation involving tangent, example 7: identify all solutions to the equation involving tangent, solving trigonometric equations in quadratic form, example 10: solving a trigonometric equation in quadratic form, example 11: solving a trigonometric equation in quadratic form by factoring, example 12: solving a trigonometric equation using algebra.

Example 13: Solving a Trigonometric Equation Quadratic in Form

Solving trigonometric equations using fundamental identities, example 14: use identities to solve an equation, example 15: solving the equation using a double-angle formula, example 16: solving an equation using an identity, solving trigonometric equations with multiple angles, example 17: solving a multiple angle trigonometric equation, solving right triangle problems, example 18: using the pythagorean theorem to model an equation.

Example 19: Using the Pythagorean Theorem to Model an Abstract Problem

Key Concepts

• When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.
• Equations involving a single trigonometric function can be solved or verified using the unit circle.
• We can also solve trigonometric equations using a graphing calculator.
• Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.
• We can also use the identities to solve trigonometric equation.
• We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval.
• Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions.

Section Exercises

Licenses & attributions, cc licensed content, specific attribution.

• Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/ [email protected] :1/Preface. License: CC BY: Attribution .

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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.

Example Question #2 : Solving Word Problems With Trigonometry

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.

Example Question #3 : Solving Word Problems With Trigonometry

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.

Example Question #4 : Solving Word Problems With Trigonometry

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.

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.

28.31 meters

23.81 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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1.4: Solving Right Triangles

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Learning Objectives

• Solve right triangles.
• Find the area of any triangle using trigonometry.
• Solve real-world problems using right triangles.
• Find the measure of an angle using inverse trig functions.

Inverse Trigonometric Ratios

In mathematics, the word inverse means “undo.” For example, addition and subtraction are inverses of each other because one undoes the other. When we use the inverse trigonometric ratios, we can find acute angle measures as long as we are given two sides.

Inverse Tangent : Labeled \(\tan ^{-1}\), the “-1” means inverse.

\(\tan ^{-1} \left(\dfrac{b}{a}\right)=m\angle B\) and \(\tan ^{-1} \left(\dfrac{a}{b}\right)=m\angle A.\)

Inverse Sine : Labeled \(\sin ^{-1}\).

\(\sin ^{-1} \left(\dfrac{b}{c}\right)=m\angle B\) and \(\sin ^{-1} \left(\dfrac{a}{c}\right)=m\angle A.\)

Inverse Cosine : Labeled \(\cos ^{-1}\).

\(\cos ^{-1} \left(\dfrac{a}{c}\right)=m\angle B\) and \(\cos ^{-1} \left(\dfrac{b}{c}\right)=m\angle A.\)

In most problems, to find the measure of the angles you will need to use your calculator. On most scientific and graphing calculators, the buttons look like \([\sin ^{-1}]\), \([\cos ^{-1}]\), and \([\(\tan ^{-1}]\). You might also have to hit a shift or 2nd button to access these functions.

Now that you know both the trig ratios and the inverse trig ratios you can solve a right triangle. To solve a right triangle, you need to find all sides and angles in it. You will usually use sine, cosine, or tangent; inverse sine, inverse cosine, or inverse tangent; or the Pythagorean Theorem .

What if you were told the tangent of \(\angle Z\) is 0.6494? How could you find the measure of \(\angle Z\)?

Example \(\PageIndex{1}\)

Solve the right triangle.

The two acute angles are congruent, making them both \(45^{\circ}\). This is a 45-45-90 triangle. You can use the trigonometric ratios or the special right triangle ratios.

Trigonometric Ratios

\(\begin{array}{rlrl} \tan 45^{\circ} & =\dfrac{15}{B C} & \sin 45^{\circ} & =\dfrac{15}{A C} \\ B C & =\dfrac{15}{\tan 45^{\circ}}=15 & A C & =\dfrac{15}{\sin 45^{\circ}} \approx 21.21 \end{array}\)

45-45-90 Triangle Ratios

\(BC=AB=15 \text{, } AC=15\sqrt{2} \approx 21.21\)

Example \(\PageIndex{2}\)

Use the sides of the triangle and your calculator to find the value of \(\angle A\). Round your answer to the nearest tenth of a degree.

In reference to \(\angle A\), we are given the opposite leg and the adjacent leg. This means we should use the tangent ratio.

\(\tan A=\dfrac{20}{25}=\dfrac{4}{5}\). So, \(\tan ^{-1} \dfrac{4}{5}=m\angle A\). Now, use your calculator.

If you are using a TI-83 or 84, the keystrokes would be: [2nd][TAN](\(\dfrac{4}{5}\)) [ENTER] and the screen looks like:

Figure \(\PageIndex{4}\)

\(m\angle A \approx 38.7^{\circ}\)

Example \(\PageIndex{3}\)

\angle A is an acute angle in a right triangle. Find \(m\angle A\) to the nearest tenth of a degree for \(\sin A=0.68\), \(\cos A=0.85\), and \(\tan A=0.34\).

\(\begin{aligned} m\angle A&=\sin ^{-1} 0.68\approx 42.8^{\circ} \\ m\angle A&=\cos ^{-1} 0.85\approx 31.8^{\circ} \\ m\angle A&=\tan ^{-1} 0.34\approx 18.8^{\circ} \end{aligned}\)

Example \(\PageIndex{4}\)

Figure \(\PageIndex{5}\)

To solve this right triangle, we need to find \(AB\), \(m\angle C\) and \(m\angle B\). Use only the values you are given.

\(\underline{AB}: \text{ Use the Pythagorean Theorem.}

\(\begin{aligned} 24^2+AB^2&=30^2 \\ 576+AB^2&=900 \\ AB^2&=324 \\ AB&=\sqrt{324}=18 \end{aligned}\)

\(\underline{m\angle B} : \text{ Use the inverse sine ratio.}\)

\(\begin{aligned} \sin B &=\dfrac{24}{30}=\dfrac{4}{5} \\ \sin ^{-1} (45) &\approx 53.1^{\circ} =m\angle B\end{aligned}\)

\(\underline{m\angle C} : \text{ Use the inverse cosine ratio.}\)

\(\cos C=\dfrac{24}{30}=\dfrac{4}{5} \rightarrow \cos ^{-1} (\dfrac{4}{5})\approx 36.9^{\circ} =m\angle C\)

Example \(\PageIndex{5}\)

When would you use sin and when would you use \(\sin ^{-1}\) ?

You would use sin when you are given an angle and you are solving for a missing side. You would use \(\sin ^{-1} \)when you are given sides and you are solving for a missing angle.

Solving the following right triangles. Find all missing sides and angles. Round any decimal answers to the nearest tenth.

Additional Resources

Interactive Element

Video: Introduction to Inverse Trigonometric Functions

Activities: Inverse Trigonometric Ratios Discussion Questions

Study Aids: Trigonometric Ratios Study Guide

Practice: Solve Right Triangles

Angles of Elevation and Depression

You can use right triangles to find distances, if you know an angle of elevation or an angle of depression .

The figure below shows each of these kinds of angles.

The angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object. For example, if you are standing on the ground looking up at the top of a mountain, you could measure the angle of elevation. The angle of depression is the angle between the horizontal line of sight and the line of sight down to an object. For example, if you were standing on top of a hill or a building, looking down at an object, you could measure the angle of depression. You can measure these angles using a clinometer or a theodolite. People tend to use clinometers or theodolites to measure the height of trees and other tall objects. Here we will solve several problems involving these angles and distances.

Finding the angle of elevation

You are standing 20 feet away from a tree, and you measure the angle of elevation to be \(38^{\circ}\). How tall is the tree?

The solution depends on your height, as you measure the angle of elevation from your line of sight. Assume that you are 5 feet tall.

The figure shows us that once we find the value of \(T\), we have to add 5 feet to this value to find the total height of the triangle. To find \(T\), we should use the tangent value:

\(\begin{aligned} tan38^{\circ}&=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{T}{20} \\ tan38^{\circ}&=\dfrac{T}{20}\\ T&=20tan38^{\circ}\approx 15.63\\ \text{Height of tree}&\approx 20.63 \text{ ft}\end{aligned}\)

You are standing on top of a building, looking at a park in the distance. The angle of depression is 53^{\circ} . If the building you are standing on is 100 feet tall, how far away is the park? Does your height matter?

Finding the angle of depression

If we ignore the height of the person, we solve the following triangle:

Given the angle of depression is \(53^{\circ}\), \(\angle A\) in the figure above is \(37^{\circ}\). We can use the tangent function to find the distance from the building to the park:

\(\begin{aligned} tan 37^{\circ}=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{d}{100}\\ tan37^{\circ} &=\dfrac{d}{100}\\d&=100 tan37^{\circ} \approx 75.36\text{ ft} \end{aligned}\)

If we take into account the height if the person, this will change the value of the adjacent side. For example, if the person is 5 feet tall, we have a different triangle:

\(\begin{aligned} tan37^{\circ}=&\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{d}{105} \\ tan37^{\circ} &=\dfrac{d}{105} \\ d&=105 tan37^{\circ} \approx 79.12\text{ ft}\end{aligned}\)

If you are only looking to estimate a distance, then you can ignore the height of the person taking the measurements. However, the height of the person will matter more in situations where the distances or lengths involved are smaller. For example, the height of the person will influence the result more in the tree height problem than in the building problem, as the tree is closer in height to the person than the building is.

Real-World Application: The Horizon

You are on a long trip through the desert. In the distance you can see mountains, and a quick measurement tells you that the angle between the mountaintop and the ground is \(13.4^{\circ}\). From your studies, you know that one way to define a mountain is as a pile of land having a height of at least 2,500 meters. If you assume the mountain is the minimum possible height, how far are you away from the center of the mountain?

\(\begin{aligned} tan 13.4^{\circ}&=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{2500}{d}\\ tan 13.4^{\circ} &=\dfrac{2500}{d} \\ d&=\dfrac{2500}{tan13.4^{\circ}} \approx 10,494 meters \end{aligned}\)

Example \(\PageIndex{6}\)

You are six feet tall and measure the angle between the horizontal and a bird in the sky to be 40^{\circ} . You can see that the shadow of the bird is directly beneath the bird, and 200 feet away from you on the ground. How high is the bird in the sky?

We can use the tangent function to find out how high the bird is in the sky:

\(\begin{aligned} tan40^{\circ} =\dfrac{height}{200} \\ height&=200 tan40^{\circ} \\ height&=(200)(.839) \\height&=167.8\end{aligned}\)

We then need to add your height to the solution for the triangle. Since you are six feet tall, the total height of the bird in the sky is 173.8 feet.

Example \(\PageIndex{7}\)

While out swimming one day you spot a coin at the bottom of the pool. The pool is ten feet deep, and the angle between the top of the water and the coin is \(15^{\circ}\). How far away is the coin from you along the bottom of the pool?

Since the distance along the bottom of the pool to the coin is the same as the distance along the top of the pool to the coin, we can use the tangent function to solve for the distance to the coin:

\(\begin{aligned} tan15^{\circ}&=\dfrac{\text{opposite}}{\text{adjacent}} \\ tan 15^{\circ} =\dfrac{10}{x} \\ x&=\dfrac{10}{tan15^{\circ}} \\ x&\approx 37.32^{\circ}\end{aligned}\)

Example \(\PageIndex{8}\)

You are hiking and come to a cliff at the edge of a ravine. In the distance you can see your campsite at the base of the cliff, on the other side of the ravine. You know that the distance across the ravine is 500 meters, and the angle between your horizontal line of sight and your campsite is \(25^{\circ}\). How high is the cliff? (Assume you are five feet tall.)

Using the information given, we can construct a solution:

\(\begin{aligned} tan 25^{\circ} =\dfrac{\text{opposite}}{\text{adjacent}}\\ tan25^{\circ}&=\dfrac{height}{500} \\ height&=\dfrac{500}{tan25^{\circ}}\\ height&=(500)(.466) \\ height&=233\text{ meters}\end{aligned}\)

This is the total height from the bottom of the ravine to your horizontal line of sight. Therefore, to get the height of the ravine, you should take away five feet for your height, which gives an answer of 228 meters.

• A 70 foot building casts an 50 foot shadow. What is the angle that the sun hits the building?
• You are standing 10 feet away from a tree, and you measure the angle of elevation to be \(65^{\circ}\). How tall is the tree? Assume you are 5 feet tall up to your eyes.
• Kaitlyn is swimming in the ocean and notices a coral reef below her. The angle of depression is \(35^{\circ}\) and the depth of the ocean, at that point is 350 feet. How far away is she from the reef?
• The angle of depression from the top of a building to the base of a car is \(60^{\circ}\). If the building is 78 ft tall, how far away is the car?

Video: Example: Determine What Trig Function Relates Specific Sides of a Right Triangle

Practice: Angles of Elevation and Depression

• Mathematics
• High School

How do you solve real-world application problems involved trigonometric ratios?? (Please help, I need answers asap!!)

Sarahplayzpiano is waiting for your help..

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Answer: find the answer in the explanation.

Step-by-step explanation:

There are various real life applications of trigonometry ratio such in building construction, oceanography, cartography, e.t.c

Its application extends to aviation industry, astronomy and oil and gas industry. Just to mention a few.

We can solve real-world applications involving trigonometric ratios by applying calculus and periodic function table and chart, and calculation involving trigonometry ratio itself ( Soh Cah Toa) e.t.c.

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