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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

  1. A1
  2. B1 2
  3. C0
  4. 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

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