Question

If $\sin (A+B)=\frac{\sqrt{3}}{2},\cos (A-B)=\frac{\sqrt{3}}{2}$ and $0<A+B\le {90}^o$,if $A>B$, then the value of $A$ and $B$ are

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

Answer 1

Answer: (A)

$A={45}^o,B={15}^o$

Let $\sin (A+B)=\frac{\sqrt{3}}{2}$
$\Rightarrow \sin (A+B)=\sin {60}^o$
$\Rightarrow A+B={60}^o$ ...(i)
Let $\cos (A-B)=\frac{\sqrt{3}}{2}$
$\Rightarrow \cos (A-B)=\cos {30}^o$
$\Rightarrow A-B={30}^o$ ...(ii)
Adding (1) and (2), we get
$2A=90$
$A={45}^{\circ }$
Subtracting (1) and (2), we have
$2B=30$
$B={15}^{\circ }$