Question

If $\tan (A+B)=1$, and $\cos (A-B)=\frac{\sqrt{3}}{2},0^o<A+B<{90}^o$ and $A>B$; find $A$ and $B$.

Answer 1

Given that

$tan(A+B)=1$ implies $(A+B)={45}^o$

$cos(A-B)=\frac{\sqrt{3}}{2}$ implies $(A-B)={30}^o$

Solving 1 and 2 we get $A={37.5}^o$ and $B={7.5}^o$