# Find all sides of a right triangle given one side and opposite angle

I've been trying to wrap my head around the following problem:

Determine the length of the side and hypotenuse in a right triangle, given the fact that one of the sides of this triangle is 16, and $$\cos \beta = \frac{1}{2}$$ where $$\beta$$ is the angle opposite to the given side.

Now since $$\cos \beta = \frac{1}{2}$$, this means that $$\beta=60^\circ$$, which in turn means that the other angle is $$30^\circ$$. Based on this, I came up with the conclusion that this triangle is a 30-60-90 triangle, and since the given side, let's denote that with $$b$$ is 16, then this means that the other side, say $$a$$, is $$16\sqrt{3}$$ and the hypotenuse is 32.

The problem is, the ratio of $$a$$ and the hypotenuse $$c$$ with these values does not seem to be $$\frac{1}{2}$$. What am I missing?

• In school we memorized the ratios as sohcahtoa. As $\sin \beta = \frac{\sqrt 3}{2}$ and Opposite $b = 16$ then $\frac{\sqrt 3}{2} = \frac{O}{H} = \frac{16}{H}$ Note that I made errors doing it in my head, got better after I wrote it down with letters. Feb 25, 2023 at 20:25

Since the side opposite to $$\beta$$ has length $$16$$, if $$a$$ is the side opposite to $$\alpha(=30^\circ)$$, then $$\frac{16}{\sin(60^\circ)}=\frac a{\sin(30^\circ)}$$, and therefore$$a=16\times\frac{\sin(30^\circ)}{\sin(60^\circ)}=\frac{16}{\sqrt3}.$$And the length of the hypotenuse is$$\sqrt{16^2+\left(\frac{16}{\sqrt3}\right)^2}=\frac{32}{\sqrt3}.$$And, indeed$$\frac{\frac{16}{\sqrt3}}{\frac{32}{\sqrt3}}=\frac12=\sin(30^\circ).$$