GATE CS 2015 Set 2 Q.8 (GA Q.8): Quantitative Aptitude

In a triangle $PQR$ , $PS$ is the angle bisector of $\angle QPR$ and $\angle QPS = 60^\circ$ . What is the length of $PS$ ?

$\dfrac{q+r}{qr}$

$\dfrac{qr}{q+r}$

$\sqrt{q^2 + r^2}$

$\dfrac{(q+r)^2}{qr}$

Official answer and solution

Solution

Express the area of triangle $PQR$ as the sum of the areas of sub-triangles $PQS$ and $PSR$ .

In $\triangle PQR$ , $PS$ bisects $\angle QPR$ , and $\angle QPS = 60^\circ$ . Because $PS$ is the angle bisector, we have:

\angle SPR = \angle QPS = 60^\circ,

which implies the total angle at $P$ is:

\angle QPR = 60^\circ + 60^\circ = 120^\circ.

The length of $PQ$ is $r$ , the length of $PR$ is $q$ , and let $x$ denote the length of the angle bisector $PS$ .

The area of $\triangle PQR$ equals the sum of areas of $\triangle PQS$ and $\triangle PSR$ :

\text{Area}(\triangle PQR) = \text{Area}(\triangle PQS) + \text{Area}(\triangle PSR)

\frac{1}{2} r q \sin(120^\circ) = \frac{1}{2} r x \sin(60^\circ) + \frac{1}{2} q x \sin(60^\circ)

Since $\sin(120^\circ) = \sin(60^\circ) = \dfrac{\sqrt{3}}{2}$ , we can divide both sides by $\dfrac{1}{2} \sin(60^\circ)$ :

r q = r x + q x = x (q + r)

x = \frac{qr}{q + r}.

Final answer

Common trap

Confusing $PS = \dfrac{qr}{q+r}$ with its reciprocal $\dfrac{q+r}{qr}$ in option A.

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