1985 AMC 12 Problem 28

Attempt Problem 28 of the 1985 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 1985 AMC 12 solutions, or check the answer key.

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

28.

In △ABC,\triangle ABC, we have ∠C=3∠A,\angle C=3\angle A, a=27a=27 and c=48.c=48. What is b?b?
3333
3535
3737
3939
not uniquely determined
Answer: B
Concepts:law of sinestrigonometric identity
Small hint:
Use the sine law to write 4827=sin⁡3Asin⁡A\frac{48}{27}=\frac{\sin3A}{\sin A}
Big hint:
After finding cos⁡A,\cos A, use B=180∘−4AB=180^\circ-4A and sin⁡4A=4sin⁡Acos⁡Acos⁡2A\sin4A=4\sin A\cos A\cos2A
Solution:
By the sine law and the triple-angle identity, 4827=sin⁡3Asin⁡A=3−4sin⁡2A. \frac{48}{27} =\frac{\sin3A}{\sin A} =3-4\sin^2A. Hence sin⁡2A=1136.\sin^2A=\frac{11}{36}. Since A<60∘,A\lt60^\circ, cos⁡A=56,\cos A=\frac{5}{6}, and therefore cos⁡2A=718.\cos2A=\frac{7}{18}. Also B=180∘−4A,B=180^\circ-4A, so another application of the sine law gives b=27sin⁡Bsin⁡A=27sin⁡4Asin⁡A=27⋅4cos⁡Acos⁡2A=35. \begin{aligned} b &=27\frac{\sin B}{\sin A}\\ &=27\frac{\sin4A}{\sin A}\\ &=27\cdot4\cos A\cos2A\\ &=35. \end{aligned} Thus the correct answer is B.
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