1972 AMC 12 Problem 27

Attempt Problem 27 of the 1972 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1972 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

If the area of △ABC\triangle ABC is 6464 square units and the geometric mean (mean proportional) between sides ABAB and ACAC is 1212 inches, then sin⁡A\sin A is equal to:
32\dfrac{\sqrt3}{2}
35\dfrac35
45\dfrac45
89\dfrac89
1517\dfrac{15}{17}
Answer: D
Concepts:triangle areatrigonometryalgebraic manipulation
Small hint:
The geometric-mean condition determines the product AB⋅ACAB\cdot AC
Big hint:
Use [△ABC]=12(AB)(AC)sin⁡A[\triangle ABC]=\frac12(AB)(AC)\sin A
Solution:
The geometric-mean condition says AB⋅AC=122=144. AB\cdot AC=12^2=144. Therefore 64=12(AB)(AC)sin⁡A=72sin⁡A, \begin{aligned} 64&=\frac12(AB)(AC)\sin A\\ &=72\sin A, \end{aligned} so sin⁡A=89.\sin A=\frac{8}{9}.
Therefore, the correct answer is D.
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