# Solving for a side in right triangles with trigonometry

CCSS Math: HSG.SRT.C.8
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 $\triangle ABC$, find $AC$.

### Solution

Step 1: Determine which trigonometric ratio to use.
Let's focus on angle $\goldD B$ since that is the angle that is explicitly given in the diagram.
Note that we are given the length of the $\purpleC{\text{hypotenuse}}$, and we are asked to find the length of the side $\blueD{\text{opposite}}$ angle $\goldD B$. The trigonometric ratio that contains both of those sides is the sine.
The trig ratios are defined for angle $A$ below.
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 $S\blueD{O}\purpleC{H}$ $C\maroonC{A}\purpleC{H}$ $T\blueD{O}\maroonC{A}$ 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( \goldD{ B}) &= \dfrac{ \blueD{\text{ opposite}} \text{ } }{\purpleC{\text{ hypotenuse} }} ~~~~~~~~\small{\gray{\text{Define sine.}}}\\\\ \sin (\goldD{50^\circ})&= \dfrac{\blueD{AC}}{\purpleC6}~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Substitute.}}} \\\\\\\\ 6\sin ({50^\circ})&= {{AC}} ~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Multiply both sides by }6.}}\\\\\\\\ 4.60&\approx AC~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Evaluate with a calculator.}}} \end{aligned}
Here's a video of Sal solving a triangle using the trigonometric ratios.

# Now let's try some practice problems.

### Problem 1

Given $\triangle DEF$, find $DE$.

Step 1: Determine which trigonometric ratio to use.
Here we are given the length of the side $\maroonC{\text{adjacent}}$ to angle $E$ and are asked to find the length of the $\purpleC{\text{hypotenuse}}$. 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 (\goldD{ E}) &= \dfrac{\maroonC{\text{ adjacent}} }{\purpleC{\text{ hypotenuse} }}~~~~~~~~\small{\gray{\text{Define cosine.}}}\\\\ \\\\ \cos (\goldD{55^\circ})&= \dfrac{\maroonC{4}}{\purpleC{ED}} ~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Substitute.}}}\\\\\\\\ ED\cdot\cos ({55^\circ})&= 4 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Multiply both sides by}ED.}}\\\\\\\\ ED&=\dfrac{4}{\cos (55^\circ)}~~~~~~~~~~~~~~\small{\gray{\text{Divide both sides by}\cos(55^\circ).}}\\\\\\\\ ED&\approx 6.97~~~~~~~~~~~~~~~~~~~~~~~\small{\gray{\text{Evaluate with a calculator.}}} \end{aligned}

### Problem 2

Given $\triangle DOG$, find $DG$.

### Problem 3

Given $\triangle TRY$, find $TY$.
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 $\sin (28^\circ)=\dfrac{20}{z}$ .
Since the acute angles of a right triangle sum to $90^\circ$, we also know that $m\angle I = 62^\circ$. 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, $\cos (62^\circ)=\dfrac{20}{z}$ .
The following two equations could be used to find $z$:
$\sin (28^\circ)=\dfrac{20}{z}$
$\cos (62^\circ)=\dfrac{20}{z}$