Question

If $P$ is a point inside $\Delta ABC$ such that $\angle PBC=\frac{A}{2},\angle PCA=\frac{B}{2}$ and$\angle PAB=\frac{C}{2}$,

then the value of $\frac{\sin (A-\frac{C}{2})\sin (B-\frac{A}{2})\sin (C-\frac{B}{2})}{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}$ is?

• A. $3$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is B $1$

Considering the above figure we get

$\frac{sin(A-\frac{C}{2})}{sin\frac{B}{2}}=\frac{PC}{PA}$

$\frac{sin(B-\frac{A}{2})}{sin\frac{C}{2}}=\frac{PA}{PB}$

$\frac{sin(C-\frac{B}{2})}{sin\frac{A}{2}}=\frac{PB}{PC}$

By multiplying all the above, we get

$\frac{sin(A-\frac{C}{2}).in(B-\frac{A}{2}).sin(C-\frac{B}{2})}{sin\frac{B}{2}.sin\frac{C}{2}.sin\frac{A}{2}}=\frac{PC}{PA}.\frac{PA}{PB}.\frac{PB}{PC}$

$=1$