1972 AHSME Problems/Problem 27

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

Draw Diagram later

We can let $AB=s$ and $AC=r$ . We can also say that the area of a triangle is $\frac{1}{2}rs\sin A$ , which we know, is $64$ . We also know that the geometric mean of $r$ and $s$ is 12, so $\sqrt{rs}=12$ .

Squaring both sides gives us that $rs=144$ . We can substitute this value into the equation we had earlier. This gives us $\frac{1}{2}\times144\times\sin A=64$ $72\sin A=64$ $\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}$

-edited for readability

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1972 AHSME Problems/Problem 27

Problem

Solution