1999 AMC 12 Problem 27

Attempt Problem 27 of the 1999 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 1999 AMC 12 solutions, or check the answer key.

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

27.

In triangle ABC,ABC, 3sin⁡A+4cos⁡B=63\sin A + 4\cos B = 6 and 4sin⁡B+3cos⁡A=1.4\sin B + 3\cos A = 1. Then ∠C\angle C in degrees is
3030
6060
9090
120120
150150
Answer: A
Concepts:trigonometric identitysystem of equations
Small hint:
Square both equations and add them
Big hint:
Use sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1 to get sin⁡(A+B),\sin(A + B), then rule out one case
Solution:
Squaring both equations and adding gives 9+16+24(sin⁡Acos⁡B+cos⁡Asin⁡B)=37, \begin{aligned} &9 + 16 \\ &\quad {}+ 24\small(\sin A \cos B + \cos A \sin B) \\ &\quad = 37, \end{aligned} so 24sin⁡(A+B)=1224\sin(A + B) = 12 and sin⁡(A+B)=12.\sin(A + B) = \tfrac12.
Then sin⁡C=sin⁡(A+B)=12,\sin C = \sin(A + B) = \tfrac12, so ∠C=30∘\angle C = 30^\circ or 150∘.150^\circ. If ∠C=150∘,\angle C = 150^\circ, then A<30∘,A \lt 30^\circ, making 3sin⁡A+4cos⁡B<6,3\sin A + 4\cos B \lt 6, a contradiction. Hence ∠C=30∘.\angle C = 30^\circ.
Thus, the correct answer is A.
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