1959 AHSME Problems/Problem 15

In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is: $\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}$

Solution

WLOG, by scaling, that the hypotenuse has length 1. Let $\theta$ be an angle opposite from some leg. Then the two legs have length $\sin\theta$ and $\cos\theta$ respectively, so we have $2\sin\theta\cos\theta = 1^2$ . From trigonometry, we know that this equation is true when $\theta = 45^{\circ}$ , so our answer is $\boxed{\textbf{(C)}}$ and we are done.

1959 AHSC (Problems • Answer Key • Resources)
Preceded by 14	Followed by Problem 16
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1959 AHSME Problems/Problem 15

Solution

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