Monoxdifly
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Suppose the angles in triangle ABC is A, B, and C. If sin A + sin B = 2 sin C, the value of $$2tan\frac12Atan\frac12B$$ is ...
A. $$\frac83$$
B. $$\sqrt6$$
C. $$\frac73$$
D. $$\frac23$$
E. $$\frac13\sqrt3$$
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 = $$\frac{sinA+sinB}{sin(A+B)}$$
$$2tan\frac12Atan\frac12B$$
$$\frac{sinA+sinB}{sin(A+B)}×tan\frac12Atan\frac12B$$
What am I supposed to do after this?
A. $$\frac83$$
B. $$\sqrt6$$
C. $$\frac73$$
D. $$\frac23$$
E. $$\frac13\sqrt3$$
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 = $$\frac{sinA+sinB}{sin(A+B)}$$
$$2tan\frac12Atan\frac12B$$
$$\frac{sinA+sinB}{sin(A+B)}×tan\frac12Atan\frac12B$$
What am I supposed to do after this?