Question

Let $p,q,r$ are lengths of an acute angled triangle $△PQR$ opposite to sides $QR,PR,PQ$ respectively. The perpendiculars are drawn from the angles $P$, $Q$ and $R$ on opposite sides and produced to meet the circumscribing circle. If these produced parts be ${\theta }_1,{\theta }_2,{\theta }_3$ respectively, then the value of $\frac{\left(p/{\theta }_1\right)+\left(q/{\theta }_2\right)+\left(r/{\theta }_3\right)}{\tan P+\tan Q+\tan R}$ is
(correct answer + 1, wrong answer - 0.25)

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$


Let $PS$ be the perpendicular from $P$ on $QR$.
When $PS$ is produced, it meets the circumscribing circle at $T$. From question, $ST={\theta }_1$

Since, angles in the same segment are equal,
$\therefore \angle PTQ=\angle PRQ=\angle R$ and $\angle PTR=\angle PQR=\angle Q$

From right-angled triangle $QST$, we get
$\tan R=\frac{QS}{ST}\cdots \left(1\right)$

From the right - angled triangle $RST$ we get
$\tan Q=\frac{SR}{ST}\cdots \left(2\right)$
Adding equations $\left(1\right)$ and $\left(2\right)$, we get
$\tan Q+\tan R=\frac{QS+SR}{ST}=\frac{QR}{ST}=\frac{p}{{\theta }_1}\cdots \left(3\right)$
Similarly, $\tan R+\tan P=\frac{q}{{\theta }_2}\cdots \left(4\right)$
and $\tan P+\tan Q=\frac{r}{{\theta }_3}\cdots \left(5\right)$

$\therefore \frac{p}{{\theta }_1}+\frac{q}{{\theta }_2}+\frac{r}{{\theta }_3}=2(\tan P+\tan Q+\tan R)$