Question

If $\sin (A-B)=\frac{1}{2}$ and $\cos (A+B)=\frac{1}{2}$, then

• A. $A={60}^0,B={45}^0$
• B. $A={30}^0,B={15}^0$
• C. $A={45}^0,B={15}^0$
• D. $A={60}^0,B={30}^0$

Answer 1

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

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

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

So $A-B=30$ & $A+B={60}^o$

Adding the two equations:

$2A={90}^o$

$\Rightarrow A={45}^o$

and $B=60-A={15}^o$

So $A={45}^o$ & $B={15}^o$.