Difference between revisions of "1987 AHSME Problems/Problem 6"

Latest revision as of 11:32, 1 March 2018

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)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 21: / Line 21: @@
 \textbf{(C)}\ 180-w-y-z \qquad \\
 \textbf{(D)}\ 2w-y-z\qquad
 \textbf{(E)}\ 180-w+y+z</math>
-\textbf{(E)}\ 50$
+== Solution ==
+By angles in a quadrilateral, <math>x = 360^{\circ} - \text{reflex } \angle ADB - y - z</math>, and by angles at a point, <math>\text{reflex } \angle ADB = 360^{\circ} - w</math>, so our expression becomes <math>360^{\circ} - (360^{\circ} - w) - y - z = w - y - z</math>, which is <math>\boxed{\text{A}}</math>.
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Difference between revisions of "1987 AHSME Problems/Problem 6"

Latest revision as of 11:32, 1 March 2018

Problem

Solution

See also