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A. \(\displaystyle \frac83\)

B. \(\displaystyle \sqrt6\)

C. \(\displaystyle \frac73\)

D. \(\displaystyle \frac23\)

E. \(\displaystyle \frac13\sqrt3\)

Since A, B, and C are the angles of triangle ABC, then C = 180° – (A + B)

sin A + sin B = 2 sin C

sin A + sin B = 2 sin(180° – (A + B))

sin A + sin B = 2 sin(A + B)

2 = \(\displaystyle \frac{sinA+sinB}{sin(A+B)}\)

\(\displaystyle 2tan\frac12Atan\frac12B\)

\(\displaystyle \frac{sinA+sinB}{sin(A+B)}×tan\frac12Atan\frac12B\)

What am I supposed to do after this?