Question

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:

$\dfrac{\sqrt3}{2}$

$\dfrac35$

$\dfrac45$

$\dfrac89$

$\dfrac{15}{17}$

Answer: D

Concepts:triangle area trigonometry algebraic manipulation

Difficulty rating: 1470

Small Hint:

The geometric-mean condition determines the product $AB\cdot AC$

Big Hint:

Use $[\triangle ABC]=\frac12(AB)(AC)\sin A$

Solution:

The geometric-mean condition says $AB\cdot AC=12^2=144.$ Therefore $\begin{aligned} 64&=\frac12(AB)(AC)\sin A\\ &=72\sin A, \end{aligned}$ so $\sin A=\frac{8}{9}.$

Therefore, the correct answer is D.

1972 AMC 12 Problem 27

Answer: D

Difficulty rating: 1470

Small Hint:

Big Hint:

Solution:

Problem 27 in Other Years