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AHSME 1999 Problem 27

    In triangle ABCABC, 3sin⁡A+4cos⁡B=63 \sin A + 4 \cos B = 6 and 4sin⁡B+3cos⁡A=14 \sin B + 3 \cos A = 1. Then ∠C\angle C in degrees is (A) 30(B) 60(C) 90(D) 120(E) 150\mathrm{(A) \ }30 \mathrm{(B) \ }60 \mathrm{(C) \ }90 \mathrm{(D) \ }120 \mathrm{(E) \ }150

    Dataset text NuminaMath-1.5. Quoted with source.

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