# 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 $\boxed{\text{A}}$.

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