Question

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

• A. ${105}^0,-{45}^0$
• B. ${45}^0,{15}^0$
• C. ${45}^0,-{15}^0$
• D. both

Answer 1

The correct option is B ${45}^0,{15}^0$

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

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

$\Rightarrow A-B={30}^{\circ }$--------------------------1

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

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

$\Rightarrow A+B={60}^{\circ }$-------------------------2

$\therefore A-B=30$

$A+B=60$

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

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

$\therefore 45-B=30$

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