Question

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

Answer 1

Given $\tan (A+B)=\sqrt{3}$$\to \left(1\right)$

and $\tan (A-B)=\frac{1}{\sqrt{3}}$$\to \left(2\right)$

and now we know that

$\tan \left(60\right)=\sqrt{3}$$\to \left(3\right)$

and

$\tan \left(30\right)=\frac{1}{\sqrt{3}}$$\to \left(4\right)$

from (1) and (3)

$A+B=60$$\to \left(5\right)$

from (2) and (4)

$A-B=30$$\to \left(6\right)$

adding (5) and (6)

$2A=90$

$\Rightarrow$$A=45$

Therefore $B=60-A=60-45=15$

Thus $A=45,B=15$