Difference between revisions of "1959 AHSME Problems/Problem 15"
Duck master (talk | contribs) (created page w/ categorization) |
(→Solution) |
||
| Line 1: | Line 1: | ||
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> | ||
| − | == Solution == | + | == Solution 1 == |
WLOG, by scaling, that the hypotenuse has length 1. Let <math>\theta</math> be an angle opposite from some leg. Then the two legs have length <math>\sin\theta</math> and <math>\cos\theta</math> respectively, so we have <math>2\sin\theta\cos\theta = 1^2</math>. From trigonometry, we know that this equation is true when <math>\theta = 45^{\circ}</math>, so our answer is <math>\boxed{\textbf{(C)}}</math> and we are done. | WLOG, by scaling, that the hypotenuse has length 1. Let <math>\theta</math> be an angle opposite from some leg. Then the two legs have length <math>\sin\theta</math> and <math>\cos\theta</math> respectively, so we have <math>2\sin\theta\cos\theta = 1^2</math>. From trigonometry, we know that this equation is true when <math>\theta = 45^{\circ}</math>, so our answer is <math>\boxed{\textbf{(C)}}</math> and we are done. | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | Look at the options. Note that if <math>\textbf{(A)}</math> is the correct answer, one of the acute angles of the triangle will measure to <math>15</math> degrees. This implies that the other acute angle of the triangle would measure to be <math>75</math> degrees, which would imply that <math>\textbf{(E)}</math> is another correct answer. However, there is only one correct answer per question, so <math>\textbf{(A)}</math> can't be a correct answer. Using a similar argument, neither <math>\textbf{(B)}</math>, <math>\textbf{(D)}</math> , nor <math>\textbf{(E)}</math> can be a correct answer. Since <math>\textbf{(C)}</math> is the only answer choice left and there must be one correct answer, the answer must be <math>\boxed{\textbf{(C)}}</math>. | ||
== See also == | == See also == | ||
Revision as of 11:23, 16 July 2020
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 14 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
