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## High school geometry

### Course: High school geometry>Unit 5

Lesson 6: 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, A, B, C, find A, C.

### Solution

Step 1: Determine which trigonometric ratio to use.
Let's focus on angle start color #e07d10, B, end color #e07d10 since that is the angle that is explicitly given in the diagram.
Note that we are given the length of the start color #aa87ff, start text, h, y, p, o, t, e, n, u, s, e, end text, end color #aa87ff, and we are asked to find the length of the side start color #11accd, start text, o, p, p, o, s, i, t, e, end text, end color #11accd angle start color #e07d10, B, end color #e07d10. The trigonometric ratio that contains both of those sides is the sine.
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}

## Now let's try some practice problems.

### Problem 1

Given triangle, D, E, F, find D, E.

### Problem 2

Given triangle, D, O, G, find D, G.

### Problem 3

Given triangle, T, R, Y, find T, Y.

### Challenge problem

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

## Want to join the conversation?

• please can someone just explain q.2 really slowly and simply? Thanks :) it's totally confusing me! •   - You are given the side OPPOSITE the 72 degree angle, which is 8.2.
- You are solving for the HYPOTENUSE.
Therefore you need the trig function that contains both the OPPOSITE and the HYPOTENUSE, which would be SINE, since sin = OPPOSITE / HYPOTENUSE.

"Let's input the value into the equation."
sin (deg) = opposite/hypotenuse
sin (72) = 8.2/DG

"Since we're solving for DG, the hypotenuse, we have to move it so that it is on the numerator. Thus, you multiply both sides of the equation by DG"
DG sin(72) = 8.2

"Again because we're solving for DG, we have to isolate DG so that it alone is on the left side of the equation. To do so, we have to move sin(72) to the other side, or in other words divide both sides of the equation by sin(72)."
DG = 8.2/sin(72)

"Now use the calculator"
8.2/sin(72) = 8.621990.....

8.62

Hope this helped :)
• why is my calculator giving me the wrong answer? • I don't understand. How angles that have simple ratios can be discovered? For example, how do you know if sin 30 is 1/2 ? • When using similar triangles, their sides are proportional. If two triangles have two congruent angles, then the triangles are similar. So, if you have a 30-60-90 triangle then the sine ratio is defined as the ratio of the length of the side opposite to the length of the hypotenuse. Doesn't matter how "big" the triangle, those sides will always have the ratio of 1/2.
• Ok I'm so lost now like I get and understand how to break it down and how to set it up to solve it but for some reason with my calculator the answers I'm getting aren't the same. For example 3*tan(37) I'm getting -2.52 yet in the comments I seen Joshua commented to Ben the break down and everything and says 3*tan(37)= 2.26 so why is my calculator showing different then everyone else? • I really need help with problem 3 • Hi! I am not really sure what you need help with but I will try to explain it the best I can.

To find TY, the side you are looking for, you need to use tan.

You use tan because of SOH CAH TOA, to use tan you use the opposite and adjacent which you have in the problem.

Look at the 37 degree. The side across from it, or the opposite, is what you are trying to find so it will be your x or unknown value. You know the adjacent side, it is three.

So you will set up your equation like this

tan(37)=x/3

The 37 comes from the degree you used as a reference point. The x comes from TOA, so you put the opposite side over the adjacent. The opposite side is x in this case and the adjacent is 3 in this case. You then find the value of tan(37) using your calculator and multiply it by three (you are using basic algebra here, treat tan(37) like a number), and you are done!

Hope that helps!
• I don't understand that how does one calculate a tangent without a calculator • ez pz lemon squeezy! • What if you have only the hypotenuse and one angle then what do you do? • can trig ratios be used on acute triangles • How do I know whether to use sin, cos, or tan? • In the example we are given right triangle 𝐴𝐵𝐶, with its right angle at 𝐶.
We are also given that angle 𝐵 measures 50°, and that the length of side 𝐴𝐵 is 6.
And we are told to find the length of side 𝐴𝐶.

The trigonometric ratios only work for the non-right angles, in this case either angle 𝐴 or angle 𝐵.
We know that the measure of angle 𝐵 is 50°, so let's use that.

– – –

In a triangle, the side opposite of an angle is the side that does not help form the angle.
In this case angle 𝐵 is formed by sides 𝐴𝐵 and 𝐵𝐶, which leaves 𝐴𝐶 as the opposite side.

In a right triangle, the hypotenuse is the side opposite of the right angle.
In this case the right angle is formed by sides 𝐴𝐶 and 𝐵𝐶, leaving 𝐴𝐵 as the hypotenuse.

In a right triangle, the side adjacent to a non-right angle is the side that together with the hypotenuse forms the angle.
We have already established that angle 𝐵 is formed by sides 𝐴𝐵 and 𝐵𝐶, and that 𝐴𝐵 is the hypotenuse.
Thereby side 𝐵𝐶 must be the adjacent side.

So, to summarize,
the measure of angle 𝐵 is 50°.
Side 𝐴𝐵, the hypotenuse, has a length of 6.
We want to find the length of side 𝐴𝐶, which is opposite of angle 𝐵.

– – –

From SOH-CAH-TOA we have that
Sine = Opposite∕Hypotenuse