Question:
If $\sin (A+B)=\frac{\sqrt{3}}{2}$ and $\cos (A-B)=\frac{\sqrt{3}}{2}, 0^{\circ}<(A+B) \leq 90^{\circ}$ and $A>B$ then find the values of $A$ and $B$.
Solution:
As we know that,
$\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
Thus,
if $\sin (A+B)=\frac{\sqrt{3}}{2}$
$\Rightarrow A+B=60^{\circ} \quad \ldots(1)$
and $\cos 30^{\circ}=\frac{\sqrt{3}}{2}$
Thus,
if $\cos (A-B)=\frac{\sqrt{3}}{2}$
$\Rightarrow A-B=30^{\circ} \quad \ldots(2)$
Solving (1) and (2), we get
$A=45^{\circ}$ and $B=15^{\circ}$
Hence, the values of $A$ and $B$ are $45^{\circ}$ and $15^{\circ}$, respectively.