# 1972 AHSME Problems/Problem 14

## Problem

A triangle has angles of $30^\circ$ and $45^\circ$. If the side opposite the $45^\circ$ angle has length $8$, then the side opposite the $30^\circ$ angle has length

$\textbf{(A) }4\qquad \textbf{(B) }4\sqrt{2}\qquad \textbf{(C) }4\sqrt{3}\qquad \textbf{(D) }4\sqrt{6}\qquad \textbf{(E) }6$

## Solution

This triangle can be split into smaller 30-60-90 and 45-45-90 triangles. The side opposite the $45^\circ$ angle has length $8,$ so the 30-60-90 triangle has sides $4, 4\sqrt3,$ and $8.$

One of the legs of the 45-45-90 triangles is $4,$ so the hypotenuse is $4\sqrt2.$ This is also the side opposite the $30^\circ$ angle, so the answer is $\textbf{(B)}.$

-edited by coolmath34

## Solution 2

Use Law of Sines to get $\frac{8}{\frac{\sqrt2}{2}}=\frac{x}{\frac{1}{2}}$. The answer is $\boxed{\textbf{(B) }4\sqrt{2}}$.

-aopspandy