# Finding the sides of a right triangle given sine and one side length

The Question: Consider a right triangle with a side of length x opposite angle A, a side of length y opposite angle B, and a hypotenuse of length z opposite the right angle. If sinB= 1/√3 and x=4, find the length of the other side (y) and the length of the hypotenuse (z).

Explanation:

This is how I sketched the problem right triangle diagram

It seems to me that the question has already gave us the value of y and z because sinB = opposite/hypotenuse = 1/√3 = y/z

When I enter 1 as the value for y and √3 as the value for z, I get the answer wrong.

What's the correct way to solve this problem? What am I not getting?

• $y/z = 1/\sqrt 3$ does not imply that $y=1, z=\sqrt 3$. You current diagram violates Pythagoreas' Theorem (and the triangle inequality); you need to account for the scaling. Nov 19 '20 at 3:45
• how do I account for scaling or what is the correct way of looking at this problem? Nov 19 '20 at 3:48
• It does violate the Pythagorean theorem Nov 19 '20 at 3:48

From $$\sin B = \frac1{\sqrt 3} = \frac yz$$ we have $$z= \sqrt 3 y$$.

By Pythagoreas' Theorem, we have $$x^2+y^2=z^2$$.

Hence:

$$16 + y^2 = (\sqrt3 y)^2 = 3y^2$$

This gives $$y^2 = 8 \leadsto y = 2\sqrt 2, z = 2\sqrt 6$$.

• Can you help me understand the first line you wrote? Nov 19 '20 at 3:53
• $\sin B = 1/\sqrt 3 = y/z$ is what you wrote. I simply multiplied both sides by $z\sqrt 3$. Nov 19 '20 at 3:56
• I don't get how you got √3y Nov 19 '20 at 4:00
• I got it, you rock! Nov 19 '20 at 4:01