Question

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

Answer 1

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

$tan(A+B)=\sqrt{3}=tan{60}^0$

$\Rightarrow A+B=60----\left(1\right)$

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

$tan(A-B)=\frac{1}{\sqrt{3}}=tan{30}^0$

$\Rightarrow A-B=30----\left(2\right)$

Adding (1) and (2),

$2A=90\Rightarrow A={45}^o$

$\therefore A+B=60\Rightarrow 45+B=60$

$\Rightarrow B=60-45={15}^o$

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