1972 AHSME Problems/Problem 27


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

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


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