Question

If $\sin (A-B)=\frac{1}{2}$ and $\cos (A+B)=\frac{1}{2}$, $0^{\circ }<A+B⩽{90}^{\circ }$, $A<B$, find A and B.

Answer 1

Given, $\sin (A-B)=\frac{1}{2}$

$\cos (A+B)=\frac{1}{2}$

$0^{\circ }<A+B\le {90}^{\circ }$ and $A<B$,

We know that, $\sin {30}^{\circ }=\frac{1}{2}$ and $\cos {60}^{\circ }=\frac{1}{2}$

Consider,

$\sin (A-B)=\frac{1}{2}$ and $\sin {30}^{\circ }=\frac{1}{2}$

$⟹(A-B)={30}^{\circ }$ ---------------(i)

Consider,

$\cos (A+B)=\frac{1}{2}$ and $\cos {60}^{\circ }=\frac{1}{2}$

$⟹(A+B)={60}^{\circ }$ ---------------(ii)

Solve (i) and (ii) :

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

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

Adding both equations,

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

$A=\frac{{90}^{\circ }}{2}={45}^{\circ }$

From (ii)

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

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

$B={60}^{\circ }-{45}^{\circ }$

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

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