Question

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

Answer 1

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

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

$\Rightarrow A+B={60}^0⟶\left(1\right)$

Also, $\tan (A-B)=1$

$\Rightarrow \tan (A-B)={\tan 45}^0$

$\Rightarrow A-B={45}^0⟶\left(2\right)$

Solving for $A$ & $B$,

$A+B={60}^0$

$\underset{–}{A-B={45}^0}$

$2A={105}^0$

$A={52.5}^0$

$\therefore$ $B={7.5}^0$