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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?

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  • $\begingroup$ 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. $\endgroup$
    – Will Jagy
    Feb 25, 2023 at 20:25

1 Answer 1

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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$$\frac{\frac{16}{\sqrt3}}{\frac{32}{\sqrt3}}=\frac12=\sin(30^\circ).$$

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