1999 AMC 12 Problem 27

Below is the professionally curated solution for Problem 27 of the 1999 AMC 12, 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.

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Concepts:trigonometric identitysystem of equations

Difficulty rating: 2120

27.

In triangle ABC,ABC, 3sinA+4cosB=63\sin A + 4\cos B = 6 and 4sinB+3cosA=1.4\sin B + 3\cos A = 1. Then C\angle C in degrees is

3030

6060

9090

120120

150150

Solution:

Squaring both equations and adding gives 9+16+24(sinAcosB+cosAsinB)=37, 9 + 16 + 24(\sin A \cos B + \cos A \sin B) = 37, so 24sin(A+B)=1224\sin(A + B) = 12 and sin(A+B)=12.\sin(A + B) = \tfrac12.

Then sinC=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 3sinA+4cosB<6,3\sin A + 4\cos B \lt 6, a contradiction. Hence C=30.\angle C = 30^\circ.

Thus, the correct answer is A.