If $sin(A+B) = cos(A-B)=\frac{\sqrt{3}}{2}$ and A and B are acute angles. The measures of angles A and B (in degrees) will be:

A = 45 and B = 15

Given :-

$sin(A+B) = cos(A-B)=\frac{\sqrt{3}}{2}$

sin(A+B) =  \(\frac{√3 }{2}\)

we know , sin 60º = \(\frac{√3 }{2}\) 

So, ( A + B ) = 60º     ------(1)

cos(A-B) =  \(\frac{√3 }{2}\)

we know , cos 30º = \(\frac{√3 }{2}\) 

So, ( A - B ) = 30º     ------(2)

Adding equation 1 and 2

2A = 90º

A = 45º

By putting value of A in equation 1

45º + B = 60º

B = 15º

ANs :- A = 45º and B = 15º