In $\triangle P Q R$, find $\Sigma(q+r) \cos P$, if $p, q, r$ denote its sides and $s=\frac{(p+q+r)}{2}$
In $\triangle P Q R$, find $\Sigma(q+r) \cos P$, if $p, q, r$ denote its sides and $s=\frac{(p+q+r)}{2}$
$s$
$s / 2$
$2 \mathrm{~s}$
$4 \mathrm{~s}$
Solution
It is given that in a $\triangle P Q R, p, q, r$ denotes its sides, so $\Sigma(q+r) \cos P$
$
\begin{array}{lr}
=q \cos P+r \cos P+r \cos Q+p \cos Q+p \cos R \\
=(q \cos P+p \cos Q)+(r \cos P+p \cos R \\
=r+q+p & +(r \cos Q+q \cos R) \\
=2 s & (\text { by projection law }) \\
& \left(\because s=\frac{p+q+r}{2}\right)
\end{array}
$