Question

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

If $B<A<{90}^{\circ }$, then,
$\angle B$ = ___

(If your answer is ${20}^{\circ }$, then type 20)

Answer 1

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

Now we know that $tan{60}^{\circ }=\sqrt{3}$

$\therefore A+B={60}^{\circ }$ ------------------- (1)

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

Now we know that $tan{30}^{\circ }=\frac{1}{\sqrt{3}}$

$\therefore A-B={30}^{\circ }$ --------------------- (2)

Adding equations 1 and 2, we get

$2A={90}^{\circ }$

$A={45}^{\circ }$

Subtracting 2 from 1, we get

$2B={30}^{\circ }$

$B={15}^{\circ }$

$\therefore \angle B$ =${15}^{\circ }$