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## Trigonometry

### Unit 1: Lesson 3

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

### 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.
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 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.
A right triangle D E F. Angle D F E is a right angle. Angle D E F is fifty degrees. Side D E is unknown. Side E F is four units.

### Problem 2

Given triangle, D, O, G, find D, G.
A right triangle D O G. Angle D O G is a right angle. Angle D G O is seventy-two degrees. Side D G is unknown. Side D O is eight point two units.

### Problem 3

Given triangle, T, R, Y, find T, Y.
A right triangle T R Y. Angle R T Y is a right angle. Angle T R Y is thirty-seven degrees. Side T Y is unknown. Side R T is three units.

### Challenge problem

In the triangle below, which of the following equations could be used to find z?
A right triangle M I X. Angle I M X is a right angle. Angle I X M is twenty-eight degrees. Side I X is unknown. Side I M is twenty units.

## 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 :)
• 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.
• why is my calculator giving me the wrong answer?
• Most calculators are set initially in Radian mode instead of degree mode. Look for a switch, and that will fix it.
• 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
• What if you have only the hypotenuse and one angle then what do you do?
• It'd depend on what are you find, the adjacent or the opposite. To find the opposite you'd use the SIN(angle)= opposite you want to find/hypotenuse you have; you'd just have to do the equation with the O/H inverse. Same to find the adjacent, but with COS.

Hope it helps you. :)
• can trig ratios be used on acute triangles
• Yes it can. Trig ratios can apply to non-right triangles. When you get to the law of sines and cosines, you will see that you can find the measures of angles and the lengths of sides on obtuse and acute triangles.
• Can someone please help me on problem 3, I would greatly appreciate it :) (given TRY, find TY)
• This problem is a tangent problem. At least, tan is the simplest for this one. I will do the working for tangent.
The given angle is 37 deg. We want to find the opposite angle, and are given the adjacent length, 3. So plug in our formula: tan*37=?/3. We multiply both sides by 3: 3*tan*37=?. You can punch that in your calculator (make sure it is in degree mode). I will not tell you the answer, but I basically did already.
Hope that was a help.
• Can someone please help me understand problem number 2? I do not know how to start.