AP EAMCET202119 Aug 2021Evening ShiftMathematicsProperties of TrianglesActual
In a triangle A B C , if 3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1 , then sin ( A + B ) is equal to
Options
- A1
- B1 2
- C0
- Dcos ⁡ C
Correct answer
B. 1 2
Step-by-step solution
Given, 3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1 On squaring and adding both the equations we get, 3 sin A + 4 cos B 2 + 4 sin B + 3 cos A 2 = 6 2 + 1 2 ⇒ 9 sin 2 A + 16 cos 2 B + 24 sin A cos B + 16 sin 2 B + 9 cos 2 A + 24 sin B cos A = 37 ⇒ 9 sin 2 A + cos 2 A + 16 cos 2 B + sin 2 B + 24 sin A cos B + sin B cos A = 37 Using sin 2 A + cos 2 A = 1 and sin ( A + B ) = sin A cos B + cos A sin B ⇒ 9 + 16 + 24 sin A + B = 37 ⇒ 24 sin A + B = 37 - 25 = 12 ⇒ sin ( A + B ) = 1 2