If sin (A + B) = √3/2 = cos (A – B), 0° < A + B ≤ 90° (A > B), find the values of A and B.
Answer :
sin (A + B) = √3/2 = cos (A – B)
sin (A + B) = √3/2
sin 60 = √3/2
sin (A + B) = sin 60°
A + B = 60° …(1)
cos (A – B) = √3/2
cos 30° = √3/2
cos (A – B) = cos 30°
A – B = 30° …(2)
By adding both the equations
A + B + A – B = 60° + 30°
2A = 90°
⇒ A = 90°/2 = 45°
Now substitute the value of A in equation (1)
45° + B = 60°
B = 60° – 45° = 15°
Hence, A = 45° and B = 15°.
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