Question

Question 3
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) = $\sqrt{3}$
$\Rightarrow tan(A+B)=tan{60}^{\circ }$
$\Rightarrow (A+B)={60}^{\circ }$ ... (i)
tan (A - B) = $\frac{1}{\sqrt{3}}$
$\Rightarrow tan(A-B)=tan{30}^{\circ }$
$\Rightarrow (A-B)={30}^{\circ }$ ... (ii)
Adding (i) and (ii); we get,
$A+B+A-B={60}^{\circ }+{30}^{\circ }$
$2A={90}^{\circ }$
$A={45}^{\circ }$
Putting the value of A in equation (i),
${45}^{\circ }+B={60}^{\circ }$
$\Rightarrow B={60}^{\circ }-{45}^{\circ }$
$\Rightarrow B={15}^{\circ }$
Thus, A = ${45}^{\circ }$ and B = ${15}^{\circ }$