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If sin(A+B) =1 and cos(A-B)= √3/2, 0°< A+B ≤ 90° and A> B, then find the measures of angles A and B.

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Solution:

sin(A+B) =1 = sin 90, so A+B = 90……………….(i) 

cos(A-B)= √3/2 = cos 30, so A-B= 30……………(ii) 

From (i) & (ii) ∠A = 60° And ∠B = 30° 

            OR 

cosθ − sin θ/ (cosθ+sin θ) = 1−√3 /(1+√3) 

Dividing the numerator and denominator of LHS by cosθ, we get 

1 − tan θ/(1+tan θ) = 1−√3 /(1+√3) 

Which on simplification (or comparison) gives tanθ = √3 Or θ= 60° 

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