90
If $\sin (A-B) = \frac{1}{2}$, $\cos (A + B) = \frac{1}{2}$; $0 < A + B \le 90^\circ$, $A > B$; find $\angle A$ and $\angle B$.
Show SolutionHide Solution↓
$$\begin{aligned}& \sin (A - B) = \sin 30^\circ \\ & A - B = 30^\circ --------(i) \\ & \cos (A+B) = \cos 60^\circ \\ & A + B = 60^\circ ---------(ii) \\ & \text{Solving (i) and (ii)} \\ & A = 45^\circ, B = 15^\circ\end{aligned}$$