Search any question & find its solution
Question:
Answered & Verified by Expert
In $\triangle P Q R$, find $\Sigma(q+r) \cos P$, if $p, q, r$ denote its sides and $s=\frac{(p+q+r)}{2}$
Options:
Solution:
2678 Upvotes
Verified Answer
The correct answer is:
$2 \mathrm{~s}$
It is given that in a $\triangle P Q R, p, q, r$ denotes its sides, so $\Sigma(q+r) \cos P$
$$
\begin{array}{lr}
=q \cos P+r \cos P+r \cos Q+p \cos Q+p \cos R \\
=(q \cos P+p \cos Q)+(r \cos P+p \cos R \\
=r+q+p & +(r \cos Q+q \cos R) \\
=2 s & (\text { by projection law }) \\
& \left(\because s=\frac{p+q+r}{2}\right)
\end{array}
$$
$$
\begin{array}{lr}
=q \cos P+r \cos P+r \cos Q+p \cos Q+p \cos R \\
=(q \cos P+p \cos Q)+(r \cos P+p \cos R \\
=r+q+p & +(r \cos Q+q \cos R) \\
=2 s & (\text { by projection law }) \\
& \left(\because s=\frac{p+q+r}{2}\right)
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.