Track / Stage 3 / 47 of 260 #47 of 2000

Problem 47

AMC 10/12, early questions

Geometry Difficulty 3.2 Find the answer

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

Your answer

Official 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

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.