1987 AHSME Problems/Problem 14

Problem

$ABCD$ is a square and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Then $\sin \theta=$

$[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,0)--(2,1)); draw((0,0)--(1,2)); label("A", (0,0), SW); label("B", (0,2), NW); label("C", (2,2), NE); label("D", (2,0), SE); label("M", (1,2), N); label("N", (2,1), E); label("$\theta$", (.5,.5), SW); [/asy]$

$\textbf{(A)}\ \frac{\sqrt{5}}{5} \qquad \textbf{(B)}\ \frac{3}{5} \qquad \textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad \textbf{(D)}\ \frac{4}{5}\qquad \textbf{(E)}\ \text{none of these}$

Solution 1

Use the Sine Area Formula. We can isolate the triangle for which the angle $\theta$ is contained in. WLOG, denote the side length of a triangle as $2$ . Our midpoints are then $1$ . Subtract the areas of the triangles that don't include the area of our desired triangle: $4-1-1-\frac{1}{2} = \frac{3}{2}.$ The Sine Area Formula tells us $\frac{1}{2}(\sqrt{5})^2\sin\theta=\frac{3}{2}.$ Solving this equation, we get $\sin\theta=\textbf{(B)}\ \frac{3}{5} \qquad$

Solution: Everyoneintexas

Solution 2

Let $\alpha = \angle MAB = \angle DAN$ , and let $s$ be the side length of $ABCD$ . Then $BM = \frac{s}{2}$ and $AM = \frac{s\sqrt{5}}{2}$ , so $\sin \alpha = \frac{1}{\sqrt{5}}$ . Therefore, $\sin \theta = \sin \left(\frac{\pi}{2} - 2\alpha \right) = \cos 2\alpha = 1 - 2 \sin^{2}\alpha = 1 - 2\cdot \frac{1}{5} = \boxed{\textbf{(B)}\ \frac{3}{5}}$ .

-j314andrews

1987 AHSME (Problems • Answer Key • Resources)
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1987 AHSME Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See also