Question

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

Answer 1

$\tan (A+B)=\sqrt{3},\tan (A+B)=\tan {60}^o$

$\Rightarrow A+B={60}^o$ .... (i)

$\tan (A+B)=\frac{1}{\sqrt{3}}$

$\tan (A-B)=\tan {30}^o$

$\Rightarrow A-B={30}^o$ ....(ii)

From (i) and (ii), we get

$B+45={60}^o$

$B={15}^o$

$\angle A={45}^o,\angle B={15}^o$