Question

If $\sin (A+B)=1$ and $\cos (A-B)=\frac{\sqrt{3}}{2}$,$0^{\circ }<A+B\le {90}^{\circ },A>B$ then find $A$ and $B$.

Answer 1

Given $\sin (A+B)=1$

$\Rightarrow \sin (A+B)=\sin {90}^o$ $(\because \sin {90}^o=1)$

$\Rightarrow A+B={90}^o$ …………$\left(1\right)$

Again, $\cos (A-B)=\sqrt{3}/2$

$\Rightarrow \cos (A-B)=\cos {30}^o$ $(\because \cos {30}^o=\sqrt{3}/2)$

$\Rightarrow A-B={30}^o$ …………$\left(2\right)$

Adding $\left(1\right)+\left(2\right)$

$A+B+A-B={90}^o+{30}^o$

$\Rightarrow 2A={120}^o$

$\Rightarrow A=120/2$

$\Rightarrow A={60}^o$

Putting $A={60}^o$ in equation-$\left(2\right)$ we get

$A-B={30}^o$

$\Rightarrow {60}^o-B={30}^o$

$\Rightarrow {60}^o-{30}^o=B$

$\Rightarrow B={30}^o$

$\therefore A={60}^o;B={30}^o$.