Question

If $\cos (A+B)=\sin (A-B)=\frac{1}{2}$, where $A$ and $B$ are positive, then smallest positive value of $A+B$ (in degrees) is :

• A. ${45}^o$
• B. ${60}^o$
• C. ${105}^o$
• D. ${150}^o$

Answer 1

The correct option is B ${60}^o$

$\cos (A+B)=\frac{1}{2}=\cos {60}^o$
$⟹A+B={60}^o$ ............. $\left(i\right)$
$\sin (A-B)=\frac{1}{2}=\sin {30}^o$
$⟹A-B={30}^o$ ............. $\left(ii\right)$
Solving eq. $\left(i\right)$ and $\left(ii\right)$ we get
$A={45}^o,B={15}^o$
$⟹A+B={60}^o$

Hence, minimum value of $A+B$ is ${60}^o$.