Question

If Sin (A+B) = 1 and Cos (A-B) =√3/2 <A+B < 90°, A > B then find A and B​

Answer 1

Answer

Given

Sin (A+B) = 1 ............(i)

Cos (A-B) =√3/2............... (ii)

From Equation (1)

Sin (A+B) = 1

=> Sin (A+B) = Sin90° ( Sin90° = 1)

=> A+B=90°............. (iii)

From Equation (2)

Cos(A-B)= √3/2

=> Cos(A-B)=Cos30°  ( Cos30°  = √3/2)

=> A-B = 30°..............(iv)

Now

From Equation (iii) and (iv)

A+B+A-B=90+30

=> 2A=120

=> A=60°

Then

A-B=30°

=> B=30°