Difference between revisions of "1959 AHSME Problems/Problem 15"
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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: <math>\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}</math> | 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: <math>\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}</math> | ||
Revision as of 12:59, 16 July 2024
Contents
Problem
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:
Solution 1
WLOG, by scaling, that the hypotenuse has length 1. Let be an angle opposite from some leg. Then the two legs have length and respectively, so we have . From trigonometry, we know that this equation is true when , so our answer is and we are done.
Solution 2
Look at the options. Note that if is the correct answer, one of the acute angles of the triangle will measure to degrees. This implies that the other acute angle of the triangle would measure to be degrees, which would imply that is another correct answer. However, there is only one correct answer per question, so can't be a correct answer. Using a similar argument, neither , , nor can be a correct answer. Since is the only answer choice left and there must be one correct answer, the answer must be .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AHSME Problems and Solutions |
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