# Difference between revisions of "1992 AIME Problems/Problem 7"

## Problem

Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$. The area of face $ABC^{}_{}$ is $120^{}_{}$, the area of face $BCD^{}_{}$ is $80^{}_{}$, and $BC=10^{}_{}$. Find the volume of the tetrahedron.

## Solution

Since the area $BCD=80=\frac{1}{2}\cdot10\cdot16$, the perpendicular from $D$ to $BC$ has length $16$.

The perpendicular from $D$ to $ABC$ is $16 \cdot \sin 30^\circ=8$. Therefore, the volume is $\frac{8\cdot120}{3}=\boxed{320}$.