# 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 1

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.

## Solution 2

Look at the options. Note that if $\textbf{(A)}$ is the correct answer, one of the acute angles of the triangle will measure to $15$ degrees. This implies that the other acute angle of the triangle would measure to be $75$ degrees, which would imply that $\textbf{(E)}$ is another correct answer. However, there is only one correct answer per question, so $\textbf{(A)}$ can't be a correct answer. Using a similar argument, neither $\textbf{(B)}$, $\textbf{(D)}$ , nor $\textbf{(E)}$ can be a correct answer. Since $\textbf{(C)}$ is the only answer choice left and there must be one correct answer, the answer must be $\boxed{\textbf{(C)}}$.

## See also

 1959 AHSC (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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