Find the angles A, B, C, of a triangle such that cos(A+B) = 1/2 and sin(B+C) = 1/2.

Answer 1

A + B = 60°; B + C = 150°. So, A + B + C + B = 210°; 180° + B = 210°; B = 30°; A = 60° - 30° = 30°;

C = 150° - 30° = 120°

Answer 2

Solution:

Since A, B, C are the angles of a triangle,

A + B + C =180° (1)

0 < A + B < 180° (2)

0 < B +C < 180° (3)

From cos(A +B ) = 1/2 and (2) it follows that

A + B = 60° (4)

From sin(B + C) = 1/2 and (3) it follows that

B + C = 30° or 150° (5)

Subtract (4) from (1) to get

C =120° (6)

Substitute (6) into (5) to get

B = 30° (7)

Substitute (7) into (4) to get

A= 30°

Answer: A = 30°, B = 30°, C = 120°

Answer 3

B + C = π / 6

A + B = π / 3

A + B + C = π / 2

Can you solve for A, B, and C and answer?