Question

Let $p,q$ and $r$ denote the lengths of the sides $QR,PR$ and $PQ$ of a triangle $PQR$ respectively. Then $p{\cos }^2\left(\frac{R}{2}\right)+r{\cos }^2\left(\frac{P}{2}\right)$

• A. equals $q$
• B. equals $\frac{p+q+r}{2}$
• C. equals $\frac{p+q+r}{4}$
• D. cannot be determined with the given data

Answer 1

The correct option is B

equals $\frac{p+q+r}{2}$

$p{\cos }^2\left(\frac{R}{2}\right)+r{\cos }^2\left(\frac{P}{2}\right)$

$p\frac{s(s-r)}{pq}+r\frac{s(s-p)}{qr}$

$=\frac{s}{q}[s-r+s-p]$

$=\frac{s}{q}[2s-r-p]$

$=\frac{s}{q}[p+q+r-r-p]$

$=s$

$=\frac{p+q+r}{2}$