Question

In $△$ ABC, $sin\left(\frac{A+B}{2}\right)=cos\left(\frac{C}{2}\right)$

• A. True
• B. False

Answer 1

The correct option is A True

In a triangle, sum of all the internal angles is 180${}^{\circ }$.
$A+B+C={180}^{\circ }$
$⟹\frac{A+B+C}{2}={90}^{\circ }$
$⟹\frac{A+B}{2}={90}^{\circ }-\frac{C}{2}$
$⟹sin\left(\frac{A+B}{2}\right)=sin({90}^{\circ }-\frac{C}{2})$
Since $sin\theta =cos({90}^{\circ }-\theta )$
$⟹sin\left(\frac{A+B}{2}\right)=cos\left(\frac{C}{2}\right)$