Solving for a side in a right triangle using the trigonometric ratios

# 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 $\triangle ABC$triangle, A, B, C, find $AC$A, C.

### Solution

**Step 1:**Determine which trigonometric ratio to use.

Let's focus on angle $\goldD B$start color goldD, B, end color goldD since that is the angle that is explicitly given in the diagram.Note that we are given the length of the $\purpleC{\text{hypotenuse}}$start color purpleC, h, y, p, o, t, e, n, u, s, e, end color purpleC, and we are asked to find the length of the side $\blueD{\text{opposite}}$start color blueD, o, p, p, o, s, i, t, e, end color blueD angle $\goldD B$start color goldD, B, end color goldD. The trigonometric ratio that contains both of those sides is thesine.The trig ratios are defined for angle $A$A below.In these definitions, it is important to understand that the terms opposite, adjacent, and hypotenuse indicate thelengthof the corresponding sides relative to the given angle, in this case, $A$A.The mnemonic device $S\blueD{O}\purpleC{H}$S, start color blueD, O, end color blueD, start color purpleC, H, end color purpleC $C\maroonC{A}\purpleC{H}$C, start color maroonC, A, end color maroonC, start color purpleC, H, end color purpleC $T\blueD{O}\maroonC{A}$T, start color blueD, O, end color blueD, start color maroonC, A, end color maroonC 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 $\triangle DEF$triangle, D, E, F, find $DE$D, E.**

*Round your answer to the nearest hundredth.*

**Your answer should be**- an integer, like $6$6
- a
*simplified proper*fraction, like $3/5$3, slash, 5 - a
*simplified improper*fraction, like $7/4$7, slash, 4 - a mixed number, like $1\ 3/4$1, space, 3, slash, 4
- an
*exact*decimal, like $0.75$0, point, 75 - a multiple of pi, like $12\ \text{pi}$12, space, p, i or $2/3\ \text{pi}$2, slash, 3, space, p, i

**Step 1:**Determine which trigonometric ratio to use.

Here we are given the length of the side $\maroonC{\text{adjacent}}$start color maroonC, a, d, j, a, c, e, n, t, end color maroonC to angle $E$E and are asked to find the length of the $\purpleC{\text{hypotenuse}}$start color purpleC, h, y, p, o, t, e, n, u, s, e, end color purpleC. The trigonometric ratio that contains both of those sides is thecosine.

**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 $\triangle DOG$triangle, D, O, G, find $DG$D, G.**

*Round your answer to the nearest hundredth.*

**Your answer should be**- an integer, like $6$6
- a
*simplified proper*fraction, like $3/5$3, slash, 5 - a
*simplified improper*fraction, like $7/4$7, slash, 4 - a mixed number, like $1\ 3/4$1, space, 3, slash, 4
- an
*exact*decimal, like $0.75$0, point, 75 - a multiple of pi, like $12\ \text{pi}$12, space, p, i or $2/3\ \text{pi}$2, slash, 3, space, p, i

### Problem 3

**Given $\triangle TRY$triangle, T, R, Y, find $TY$T, Y.**

*Round your answer to the nearest hundredth.*

**Your answer should be**- an integer, like $6$6
- a
*simplified proper*fraction, like $3/5$3, slash, 5 - a
*simplified improper*fraction, like $7/4$7, slash, 4 - a mixed number, like $1\ 3/4$1, space, 3, slash, 4
- an
*exact*decimal, like $0.75$0, point, 75 - a multiple of pi, like $12\ \text{pi}$12, space, p, i or $2/3\ \text{pi}$2, slash, 3, space, p, i

# Challenge problem

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

We are given the length of the side opposite angle $X$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 $\sin (28^\circ)=\dfrac{20}{z}$sine, left parenthesis, 28, degree, right parenthesis, equals, start fraction, 20, divided by, z, end fraction .Since the acute angles of a right triangle sum to $90^\circ$90, degree, we also know that $m\angle I = 62^\circ$m, angle, I, equals, 62, degree. 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, $\cos (62^\circ)=\dfrac{20}{z}$cosine, left parenthesis, 62, degree, right parenthesis, equals, start fraction, 20, divided by, z, end fraction .The following two equations could be used to find $z$z:

$\sin (28^\circ)=\dfrac{20}{z}$sine, left parenthesis, 28, degree, right parenthesis, equals, start fraction, 20, divided by, z, end fraction$\cos (62^\circ)=\dfrac{20}{z}$cosine, left parenthesis, 62, degree, right parenthesis, equals, start fraction, 20, divided by, z, end fraction

Solving for a side in a right triangle using the trigonometric ratios