Question

If A, B and C are interior angles of a triangle ABC, and
$\sin \left(\frac{B+C}{2}\right)=\cos \left(x\right),$
then find thevalue of x.

• A. $\frac{B}{2}$
• B. $\frac{A}{2}$
• C. $2A$
• D. $A$

Answer 1

The correct option is B $\frac{A}{2}$

Given A,B and C are interior angles of a triangle,

$\therefore A+B+C={180}^{\circ }...\left(i\right)$

Dividing both sides by 2, we get

$\frac{A}{2}+\frac{B+C}{2}={90}^{\circ }$

$\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}...\left(ii\right)$

Taking sin ratio on both sides,

$\Rightarrow \sin \left(\frac{B+C}{2}\right)=\sin ({90}^{\circ }-\frac{A}{2})...\left(iii\right)$

$\Rightarrow \sin \left(\frac{B+C}{2}\right)=\cos \left(\frac{A}{2}\right)...\left(iv\right)$

Comparing
$\sin \left(\frac{B+C}{2}\right)=\cos \left(x\right)$
with
$\sin \left(\frac{B+C}{2}\right)=\cos \left(\frac{A}{2}\right)$,
we get

$\Rightarrow x=\frac{A}{2}$