Solving for a side in right triangles with trigonometry

Learn how to use trig functions to find an unknown side length in a right triangle.

We can use trig ratios to find unknown sides in right triangles.

Let's look at an example.

Given

△ A B C

, find

A C

Solution

Step 1: Determine which trigonometric ratio to use.

Let's focus on angle $B$ ‍ since that is the angle that is explicitly given in the diagram.

A right triangle A B C. Angle A C B is a right angle. Angle A B C is fifty degrees and is highlighted. Side A C is unknown. Side A B is six units.

Note that we are given the length of the $hypotenuse$ ‍, and we are asked to find the length of the side $opposite$ ‍ angle $B$ ‍. The trigonometric ratio that contains both of those sides is the sine.
The trig ratios are defined for angle $A$ ‍ below.
A right triangle A B C where B A C is the angle of reference. Angle A C B is a right angle. Side A B is the hypotenuse. Side B C is the opposite. Side A C is the A C.
In these definitions, it is important to understand that the terms opposite, adjacent, and hypotenuse indicate the length of the corresponding sides relative to the given angle, in this case, $A$ ‍.
The mnemonic device $S O H$ ‍ $C A H$ ‍ $T O A$ ‍ can help us remember these definitions.

Step 2: Create an equation using the trig ratio sine and solve for the unknown side.

$\begin{aligned} \sin (B) & = \frac{opposite}{hypotenuse} Define sine. \\ \sin (50^{\circ}) & = \frac{A C}{6} Substitute. \\ 6 \sin (50^{\circ}) & = A C Multiply both sides by 6. \\ 4.60 & \approx A C Evaluate with a calculator. \end{aligned}$ ‍

Now let's try some practice problems.

Problem 1

Given $△ D E F$ ‍, find $D E$ ‍.
Round your answer to the nearest hundredth.

Step 1: Determine which trigonometric ratio to use.

Here we are given the length of the side $adjacent$ ‍ to angle $E$ ‍ and are asked to find the length of the $hypotenuse$ ‍. The trigonometric ratio that contains both of those sides is the cosine.

Step 2: Create an equation using the trig ratio cosine and solve for the unknown side.

$\begin{aligned} \cos (E) & = \frac{adjacent}{hypotenuse} Define cosine. \\ \cos (55^{\circ}) & = \frac{4}{E D} Substitute. \\ E D \cdot \cos (55^{\circ}) & = 4 Multiply both sides by E D . \\ E D & = \frac{4}{\cos (55^{\circ})} Divide both sides by \cos (55^{\circ}) . \\ E D & \approx 6.97 Evaluate with a calculator. \end{aligned}$ ‍

Problem 2

Given $△ D O G$ ‍, find $D G$ ‍.
Round your answer to the nearest hundredth.

Problem 3

Given $△ T R Y$ ‍, find $T Y$ ‍.
Round your answer to the nearest hundredth.

Challenge problem

In the triangle below, which of the following equations could be used to find $z$ ‍?

We are given the length of the side opposite angle

X

, and we are asked to find the length of the hypotenuse. The trigonometric ratio that contains both of those sides is the sine. So

\sin (28^{\circ}) = \frac{20}{z}

Since the acute angles of a right triangle sum to

90^{\circ}

, we also know that

m ∠ I = 62^{\circ}

. In relation to this angle, we are given the length of the adjacent side and are asked to find the length of the hypotenuse. The trigonometric ratio that contains both of those sides is the cosine. So,

\cos (62^{\circ}) = \frac{20}{z}

The following two equations could be used to find

z

$\sin (28^{\circ}) = \frac{20}{z}$ ‍

$\cos (62^{\circ}) = \frac{20}{z}$ ‍

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