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8. Right Triangles & Trigonometry
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Geometry

8.04 Solving right triangles

Lesson
Practice

Solving for sides in a right triangle

The trigonometric ratios sine, cosine, and tangent each relate an angle to a pair of sides in a right triangle. If we know an acute angle measure in the right triangle and a side length, we can solve for another side length of the triangle using sine, cosine, or tangent as follows:

  1. Highlight the reference angle.
  2. Identify which sides are the hypotenuse, opposite, and adjacent - label if desired.
  3. Determine which trigonometric ratio to use. Choose the ratio for which we have one of the required sides given and one is the missing side:
    • Sine: opposite and hypotenuse
    • Cosine: adjacent and hypotenuse
    • Tangent: opposite and adjacent
  4. Set up the chosen ratio: \sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}
  5. Solve for the variable using inverse operations.

Remember the following:

  • For a given angle \theta, the value of each trigonometric ratio stays the same no matter the size of the triangle
  • Your calculator must be set to degree mode when working with degrees in trigonometric ratios

Examples

Example 1

Find the missing side length for each triangle. Round your answer to two decimal places.

a
A right teiangle with a base of length f inches and hypotenuse of 8 inches. The angle adjacent the side of length f inches measures 47 degrees.
Worked Solution
Create a strategy

We want to start by looking at the figure and determine how the labeled side lengths and angles are related. The triangle is a right triangle. The hypotenuse side length is 8 \text{ in}. With respect to the 47 \degree angle, f is the adjacent side.

We can use this information to write the appropriate trigonometric ratio and then solve for f.

Apply the idea
\displaystyle \cos\theta\displaystyle =\displaystyle \dfrac{\text{adjacent}}{\text{hypotenuse}}
\displaystyle \cos(47\degree)\displaystyle =\displaystyle \dfrac{f}{8}Substitution
\displaystyle 8\cos(47\degree)\displaystyle =\displaystyle fMultiply both sides of the equation by 8
\displaystyle 5.46\displaystyle =\displaystyle fEvaluate the multiplication and round to two decimal places
Reflect and check

The answer for f should be less than the hypotenuse, since the hypotenuse is the longest side of the triangle. This is a good check when solving for missing side lengths.

b
A right triangle with a hypotenuse of length g feet and a base of length 11 feet. Opposite the base is angle measuring 42 degrees.
Worked Solution
Create a strategy

We want to look at the diagram and determine how the given information is related. The hypotenuse is g since the side is opposite the right angle. The side with length of 11 \text{ ft} is the opposite side with respect to the labeled angle of 42 \degree.

We can use this information to write the appropriate trigonometric ratio and then solve for g.

Apply the idea
\displaystyle \sin \theta\displaystyle =\displaystyle \dfrac{\text{opposite}}{\text{hypotenuse}}
\displaystyle \sin(42 \degree)\displaystyle =\displaystyle \dfrac{11}{g}Substitution
\displaystyle g\sin(42 \degree)\displaystyle =\displaystyle 11Multiply both sides of the equation by g
\displaystyle g\displaystyle =\displaystyle \dfrac{11}{\sin(42 \degree)}Divide both sides of the equation by \sin(42 \degree)
\displaystyle g\displaystyle =\displaystyle 16.44Evaluate the division and round to two decimal places
Reflect and check

Suppose we wanted to find the length of the third unknown side. We could use the Pythagorean theorem or evaluate tangent of the reference angle, as shown:

\displaystyle a^2+b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle 11^2+?^2\displaystyle =\displaystyle 16.44^2Substitution
\displaystyle 121 + ?^2\displaystyle =\displaystyle 270.27Evaluate the exponents
\displaystyle ?^2\displaystyle =\displaystyle 149.27Subtract 121 from both sides of the equation
\displaystyle ?\displaystyle =\displaystyle 12.22Evaluate the square root of both sides of the equation
\displaystyle \tan \theta\displaystyle =\displaystyle \dfrac{\text{opposite}}{\text{adjacent}}
\displaystyle \tan(42 \degree)\displaystyle =\displaystyle \dfrac{11}{?}Substitution
\displaystyle ? \cdot \tan(42 \degree)\displaystyle =\displaystyle 11Multiply both sides by ?
\displaystyle ?\displaystyle =\displaystyle 12.22Divide both sides by \tan(42 \degree)

Using either approach, the third unknown side of the triangle is 12.22 \text{ ft}.

Example 2

A girl is flying a kite that is attached to the end of a 23.4 \text{ m} length of string. The angle between the string and the vertical is 21 \degree. If the girl is holding the string 2.1 \text{ m} above the ground, find the vertical distance from the ground to the place where the string attaches to the kite. Round your answer to two decimal places.

A girl is flying a kite that is attached to the end of a 23.4 meter length of string. The angle between the string and the vertical is 21 degrees. The girl is holding the string 2.1 meters above the ground
Worked Solution
Create a strategy

Let's call the vertical distance from the ground to the kite h. If we could find the length of the vertical leg of the triangle, we calculate h. Let's call the length of the vertical leg of the triangle x.

A girl is flying a kite that is attached to the end of a 23.4 meter length of string. The angle between the string and the vertical is 21 degrees. The girl is holding the string 2.1 meters above the ground. Ask your techer for more information

First, calculate an acute angle in the right triangle to use as a reference angle, then use a trigonometric ratio to find the value of x. Then, we can calculate the vertical distance from the kite to the ground.

Apply the idea

Since the angle between the string and the horizontal is complementary to 21 \degree, the reference angle in the triangle must be 69 \degree.

\displaystyle \sin \left( \theta \right)\displaystyle =\displaystyle \dfrac{\text{opposite}}{\text{hypotenuse}}
\displaystyle \sin \left( 69 \degree \right)\displaystyle =\displaystyle \dfrac{x}{23.4}Substitution
\displaystyle 23.4 \sin \left( 69 \degree \right)\displaystyle =\displaystyle xMultiply both sides of the equation by 23.4
\displaystyle 21.85\displaystyle =\displaystyle xEvaluate the multiplication

The vertical distance, h, from the ground to the kite is the sum of the distance from the ground to where the string starts and the vertical distance, x, from the string to the kite. That means h=2.1+x, where x=21.85. So, h=2.1+21.85=23.95. The kite is 23.95 \text{ m} from the ground.

Example 3

Find the value of x, the side length of the parallelogram, rounding your answer to the nearest inch.

A parallelogram with an acute angle measuring 52 degrees is cut by a dashed segment to form a right triangle and a trapezoid.The hypotenuse of the right triangle measures x inches.The trapezoid's base are 32 inches and 11 inches.
Worked Solution
Create a strategy

We can use what we know about parallelograms to label and find missing pieces in order to fill out some information in the right triangle with the hypotenuse of x \text{ in}.

Since opposite angles in a parallelogram are congruent, we can label the opposite angle of 52 \degree as 52 \degree. Also, since opposite sides of a parallelogram are congruent, we know that the length of the bottom of the parallelogram must total 32 \text{ in}, so the remaining length is 21 \text{ in}:

A parallelogram with an acute angle measuring 52 degrees is cut by a dashed segment from the vertex of the obtuse angle to form a right triangle and a trapezoid.The hypotenuse of the right triangle measures x inches.The trapezoid's base are 32 inches and 11 inches. One of the sides of the right triangle mesures 21 inches.

The side length adjacent to the reference angle in the small right triangle of the parallelogram is given, and we will need to solve for hypotenuse x, so we can use cosine of the reference angle to solve.

Apply the idea
\displaystyle \cos \left( \theta \right)\displaystyle =\displaystyle \dfrac{\text{adjacent}}{\text{hypotenuse}}
\displaystyle \cos \left( 52 \degree \right)\displaystyle =\displaystyle \dfrac{21}{x}Substitution
\displaystyle x \cdot \cos \left( 52 \degree \right)\displaystyle =\displaystyle 21Multiply both sides of the equation by x
\displaystyle x\displaystyle =\displaystyle \dfrac{21}{\cos \left( 52 \degree \right)}Divide both sides of the equation by \cos \left( 52 \degree \right)
\displaystyle x\displaystyle =\displaystyle 34Evaluate the division and round to the nearest whole number
Reflect and check

Note that the diagram is not drawn to scale, so the length of the top of the parallelogram, while shorter than the side of the parallelogram, may look longer.

Idea summary

If we know an acute angle measure in the right triangle and a side length, we can solve for another side length of the triangle using sine, cosine, or tangent:

  1. Highlight the reference angle.
  2. Identify which sides are the hypotenuse, opposite, and adjacent - label if desired.
  3. Determine which trigonometric ratio to use. Choose the ratio for which we have one of the required sides given and one is the missing side:
    • Sine: opposite and hypotenuse
    • Cosine: adjacent and hypotenuse
    • Tangent: opposite and adjacent
  4. Set up the chosen ratio: \sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}
  5. Solve for the variable using inverse operations.

Solving for angles in a right triangle

The trigonometric ratios take angles and give side ratios, but we are going to need functions that take side ratios and give angles.

We are already familiar with certain operations having inverse operations that "undo" them - such as addition and subtraction, or multiplication and division. In a similar manner, we can apply inverse trigonometric functions to reverse, or "undo", the existing trigonometric functions.

The functions sine, cosine, and tangent each take an angle measure as input and return a side ratio as output.

Inverse trigonometric functions

Functions that determine the measure of an acute angle of a right triangle, given a ratio between two of its sides

For a given angle \theta we have:\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}

Two right triangles showing 2 cases on how to find the opposite side of angle theta, the adjacent side and the hypotenuse. Ask your teacher for more information

The inverse functions work in reverse, taking in the ratio of two known sides as the input and returning a missing angle measure as the output.

So the three inverse trigonometric functions are: \sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)= \theta \qquad \cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) = \theta \qquad \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right) = \theta

We can use these inverse trigonometric functions to determine an unknown angle measure when we know a pair of side lengths of a right-triangle.

As these inverse functions "undo" the original functions, we have\sin^{-1}\left(\sin\left(x\right)\right) = x \qquad \cos^{-1}\left(\cos\left(x\right)\right) = x \qquad \tan^{-1}\left(\tan\left(x\right)\right) = xand\sin\left(\sin^{-1}\left(x\right)\right) = x \qquad \cos\left(\cos^{-1}\left(x\right)\right) = x \qquad \tan\left(\tan^{-1}\left(x\right)\right) = x

Two input machines showing that if the input is theta which passes through the sine machine,the output becomes sin theta. When the sin theta becomes the input theta of the inverse of sine theta machine, the putput is theta.

It is important to note that while the notation used to represent these inverse trigonometric functions looks like a power of -1, these functions are not the same as the reciprocals of the trigonometric functions. That is:\sin^{-1}(x) \neq \frac{1}{\sin(x)} \qquad \cos^{-1}(x) \neq \frac{1}{\cos(x)} \qquad \tan^{-1}(x) \neq \frac{1}{\tan(x)}

Exploration

Explore the applet by dragging the hypotenuse of the right triangle.

Loading interactive...
  1. Change the hypotenuse to match the desired ratio. Calculate the inverse trigonometric ratio, then check the box to show the angle. Does your calculated measurement match the actual angle measure?
  2. Repeat by clicking New Goal.

Examples

Example 4

If \cos \theta=0.256, find \theta. Round your answer to two decimal places.

Worked Solution
Create a strategy

We want to undo the cosine function, that is take the inverse trigonometric functions of both sides, to solve for \theta.

Apply the idea
\displaystyle \cos \theta\displaystyle =\displaystyle 0.256
\displaystyle \cos^{-1}\left(\cos\left(\theta\right)\right) \displaystyle =\displaystyle \cos^{-1}\left(0.256\right)Apply the inverse function to both sides
\displaystyle \theta\displaystyle =\displaystyle \cos^{-1}\left(0.256\right)Evaluate the inverse function on the left side of the equation
\displaystyle \theta\displaystyle =\displaystyle 75.17 \degreeEvaluate the inverse trigonometric function
Reflect and check

We need to use a calculator here to evaluate cosine inverse. Make sure your calculator is in degree mode when working with inverse trigonometric ratios.

Example 5

Solve for the value of x. Round your answer to two decimal places.

A right triangle with an angle labeled x degrees. Opposite this angle is side 20 centimeters long. Adjacent to this angle is side 25 centimeters long.
Worked Solution
Create a strategy

With respect to x, we are given the opposite side and adjacent side of the right triangle. The opposite side has a length of 20 \text{ cm} and the adjacent side has a length of 25 \text{ cm}. Use this information to set up the appropriate triogonometric ratio and use the correct inverse trigonometric ratio to solve for x.

Apply the idea
\displaystyle \tan x \displaystyle =\displaystyle \dfrac{20}{25}
\displaystyle \tan^{-1}\left(\tan\left(x \right)\right)\displaystyle =\displaystyle \tan^{-1}\left(\dfrac{20}{25}\right)Apply the inverse function to both sides
\displaystyle x \displaystyle =\displaystyle \tan^{-1}\left(\dfrac{20}{25}\right)Evaluate the inverse function on the left side of the equation
\displaystyle x \displaystyle =\displaystyle 38.66 Evaluate the inverse trigonometric function

Example 6

A ladder of length 3.46 \text{ m} needs to make an angle with the wall of between 10 \degree and 20 \degree for it to be safe to stand on. At steeper angles, the ladder is at risk of toppling backwards when the climber leans away from it and at shallower angles, the ladder may lose its grip on the ground.

A ladder leaning on a wall.

The base of the ladder is placed at 190 \text{ cm}, 42 \text{ cm}, or 86 \text{ cm} from the wall. At which of these base lengths is the ladder safest to climb? Explain how you found your answer.

Worked Solution
Create a strategy

We know the ladder is safest to climb when the ladder makes an angle with the wall between 10 \degree and 20 \degree. So, let's call the angle the ladder makes with the wall x \degree. We can also see that the ladder, wall, and base length from wall to ladder makes a right triangle, so we can use trigonometric ratios to help find x.

We can label the diagram with the given information for each scenario, then decide which trigonometric ratio will be most useful in helping us find the missing angle measure since the ladder creates a right triangle when leaning against the wall.

Since the distance from the base of the ladder to the wall is given in centimeters and the ladder is given in meters, we will first convert the distances to meters:\begin{aligned} 190 \text{ cm} \cdot \dfrac{ 1 \text{ m}}{100 \text{ cm}}= \dfrac{190}{100} \text{ m} = 1.9 \text{ m} \\ 42 \text{ cm} \cdot \dfrac{ 1 \text{ m}}{100 \text{ cm}}= \dfrac{42}{100} \text{ m} = 0.42 \text{ m} \\ 86 \text{ cm} \cdot \dfrac{ 1 \text{ m}}{100 \text{ cm}}= \dfrac{86}{100} \text{ m} = 0.86 \text{ m} \end{aligned}

Apply the idea

First, draw the diagram with a 1.9 \text{ m} distance from the base of the ladder to the wall. Since the opposite side of the reference angle and the hypotenuse of the triangle are given, we will use the sine function.

A ladder leaning against the wall creating x degrees with the wall. the base of the ladder is 1.9 meters away from the wall. The ladder is 3.46 meters
\displaystyle \sin x \degree\displaystyle =\displaystyle \dfrac{1.9}{3.46}
\displaystyle \sin^{-1}\left(\sin\left(x \degree\right)\right)\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{1.9}{3.46}\right)Apply the inverse function to both sides
\displaystyle x \degree\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{1.9}{3.46}\right)Evaluate the inverse function on the left side of the equation
\displaystyle x \degree\displaystyle =\displaystyle 33.3 \degreeEvaluate the inverse trigonometric function

Next, draw the diagram with a 0.42 \text{ m} distance from the base of the ladder to the wall. Set up the trigonometric ratio and solve for x.

A ladder leaning against the wall creating x degrees with the wall. the base of the ladder is 0.42 meters away from the wall. The ladder is 3.46 meters
\displaystyle \sin x \degree\displaystyle =\displaystyle \dfrac{0.42}{3.46}
\displaystyle \sin^{-1}\left(\sin\left(x \degree\right)\right)\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{0.42}{3.46}\right)Apply the inverse function to both sides
\displaystyle x \degree\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{0.42}{3.46}\right)Evaluate the inverse function on the left side of the equation
\displaystyle x \degree\displaystyle =\displaystyle 7 \degreeEvaluate the inverse trigonometric function

Finally, draw the diagram with a 0.86 \text{ m} distance from the base of the ladder to the wall. Set up the trigonometric ratio and solve for x.

A ladder leaning against the wall creating x degrees with the wall. the base of the ladder is 0.86 meters away from the wall. The ladder is 3.46 meters
\displaystyle \sin x \degree\displaystyle =\displaystyle \dfrac{0.86}{3.46}
\displaystyle \sin^{-1}\left(\sin\left(x \degree\right)\right)\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{0.86}{3.46}\right)Apply the inverse function to both sides
\displaystyle x \degree\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{0.86}{3.46}\right)Evaluate the inverse function on the left side of the equation
\displaystyle x \degree\displaystyle =\displaystyle 14.4 \degreeEvaluate the inverse trigonometric function

The distance that leads to an angle with the wall of between 10 \degree and 20 \degree is when the ladder is placed 86 \text{ cm} from the wall. This forms a safe 14.4 \degree angle between the ladder and the wall.

Example 7

An isosceles triangle has equal side lengths of 10 \text{ in} and a base of 8 \text{ in} as shown.

Isosceles triangle A B C. Sides A C and C B have single tick mark which both measures 10 inches. Side A B measures 8 inches

Find m\angle A, to the nearest tenth of a degree.

Worked Solution
Create a strategy

Construct an altitude from C to \overline{AB}. Since the altitude of an isosceles triangle is also its median, the segment constructed is also a perpendicular bisector of \overline{AB}, so the length of \overline{AB} is cut in half and we have two right triangles.

Isosceles triangle A B C. Sides A C and C B have single tick mark which both measures 10 inches. A segment which is the perpendicular bisector divides sides A B into 4 inches parts

The reference angle \angle A in the newly construct right triangle is shown with its hypotenuse and adjacent side. We can use the cosine function to find the unknown angle.

Apply the idea
\displaystyle \cos \left( m \angle A \degree \right)\displaystyle =\displaystyle \dfrac{4}{10}
\displaystyle \cos^{-1}\left(\cos\left(m \angle A \degree\right)\right)\displaystyle =\displaystyle \cos^{-1}\left(\dfrac{4}{10}\right)Apply the inverse function to both sides
\displaystyle m \angle A \degree\displaystyle =\displaystyle \cos^{-1}\left(\dfrac{4}{10}\right)Evaluate the inverse function on the left side of the equation
\displaystyle m \angle A \degree\displaystyle =\displaystyle 66.4 \degreeEvaluate the inverse trigonometric function
Idea summary

We can use these inverse trigonometric functions to determine an unknown angle measure when we know a pair of side lengths of a right-triangle: \sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)= \theta \qquad \cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) = \theta \qquad \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right) = \theta

Outcomes

G.TR.4

The student will model and solve problems, including those in context, involving trigonometry in right triangles and applications of the Pythagorean Theorem.

G.TR.4b

Find and verify trigonometric ratios using right triangles.

G.TR.4c

Model and solve problems, including those in context, involving right triangle trigonometry (sine, cosine, and tangent ratios).

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