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Difference between revisions of "1972 AHSME Problems/Problem 27"

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Revision as of 01:53, 13 January 2018

Problem

If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to

$\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad \textbf{(B) }\frac{3}{5}\qquad \textbf{(C) }\frac{4}{5}\qquad \textbf{(D) }\frac{8}{9}\qquad \textbf{(E) }\frac{15}{17}$

Solution

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