Question

$\sin (A+B)=\frac{\surd 3}{2}$ and $\cos (A-B)=\frac{\surd 3}{2}$, where $(A+B)$ and $(A-B)$ are acute angles. Find the values of $A$ and $B$ $A,B<{90}^0$

Answer 1

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

$\Rightarrow A+B=\frac{\pi }{3}$ .......$\left(1\right)$

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

$\Rightarrow A-B=\frac{\pi }{6}$ ......$\left(2\right)$

Adding eqns$\left(1\right)$ and $\left(2\right)$ we get

$A+B+A-B=\frac{\pi }{3}+\frac{\pi }{6}$

$\Rightarrow 2A=\frac{2\pi +\pi }{6}=\frac{3\pi }{6}=\frac{\pi }{2}$

$\therefore A=\frac{\pi }{4}$

Substitute $A=\frac{\pi }{4}$ in $\left(1\right)$ we get

$A+B=\frac{\pi }{3}$

$\Rightarrow B=\frac{\pi }{3}-A$

$\Rightarrow B=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$

$\therefore A=\frac{\pi }{4}$ and $B=\frac{\pi }{12}$