# 1987 AHSME Problems/Problem 6

## Problem

In the $\triangle ABC$ shown, $D$ is some interior point, and $x, y, z, w$ are the measures of angles in degrees. Solve for $x$ in terms of $y, z$ and $w$.

$[asy] draw((0,0)--(10,0)--(2,7)--cycle); draw((0,0)--(4,3)--(10,0)); label("A", (0,0), SW); label("B", (10,0), SE); label("C", (2,7), W); label("D", (4,3), N); label("x", (2.25,6)); label("y", (1.5,2), SW); label("z", (7.88,1.5)); label("w", (4,2.85), S); [/asy]$

$\textbf{(A)}\ w-y-z \qquad \textbf{(B)}\ w-2y-2z \qquad \textbf{(C)}\ 180-w-y-z \qquad \\ \textbf{(D)}\ 2w-y-z\qquad \textbf{(E)}\ 180-w+y+z$

## Solution

By angles in a quadrilateral, $x = 360^{\circ} - \text{reflex } \angle ADB - y - z$, and by angles at a point, $\text{reflex } \angle ADB = 360^{\circ} - w$, so our expression becomes $360^{\circ} - (360^{\circ} - w) - y - z = w - y - z$, which is $\boxed{\text{A}}$.

 1987 AHSME (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions