GATE CSE 2015 | Set 2 | GA | Question: 8

Question

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

• A. $\left(\dfrac{(q+r)} {qr}\right)$
• B. $\left(\dfrac {qr} {q+r}\right)$
• C. $\large \sqrt {(q^2 + r^2)}$
• D. $\left(\dfrac{(q+r)^2} {qr}\right)$

Comments

It is just a coincidence that your approach gives right answer here but it doesn't work everywhere. If something is not mentioned then you can only assume default cases not whatever you want. It is not always true that angle bisector is perpendicular so you should not assume it.

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Angle bisector is not always the perpendicular bisector, however if we have options with us, we can take a specific case and see which options do not match, so as to eliminate incorrect ones..

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How <QPR= <QPS=60°

Answer 1

Area of a $\triangle=\dfrac{1}{2}\times ac \sin B =\dfrac{1}{2}\times bc \sin A =\dfrac{1}{2} ab \sin C$

so, Here area $(\triangle PQR)= \text{area } (\triangle PQS) + \text{area }(\triangle PRS)$

$\dfrac{1}{2}\; rq \sin 120 =\dfrac{1}{2} PS\times r \sin 60 +\dfrac{1}{2} PS\times q \sin 60$

$\Rightarrow PS =\dfrac{rq}{(r+q)}$
so, choice (B) is correct..

Comments

how     1/2 rq sin120 ?

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Angle bisector divides angle in two equal half.If one half is 60 other is also 60 ,making it 120

Answer 2

As per Angle Bisector theorem,

$\qquad\;\;\dfrac{QS}{SR} =\dfrac{r}{q}$
$\dfrac{QS}{(p-QS)} =\dfrac{r}{q}$
$\qquad \quad QS =\dfrac{pr}{(q+r)} \quad \quad \longrightarrow(1)$

We have in a triangle $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
 

So, from $\triangle QPS,\;\; \dfrac{QS}{\sin 60} =\dfrac{PS}{\sin Q}$

$\qquad PS =\dfrac{QS\times \sin Q}{\sin 60}\quad \quad\longrightarrow (2)$
 

From $\triangle PQR,\;\; \dfrac{p}{\sin 120} =\dfrac{q}{\sin Q}$

$p =\dfrac{q\times \sin 120}{\sin Q} =\dfrac{q\times \sin 60}{\sin Q}\quad \quad\longrightarrow (3)$

So, from $(1), (2)$ and $(3),$

$PS =\dfrac{qr}{(q+r)}$

B choice.

http://en.wikipedia.org/wiki/Angle_bisector_theorem

Comments

yes @Arjun

Please do and equation 2 and 3 as well

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How is q*Sin120 = Sin60 ( in equation 3)

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sin 120 = sin (180 – 60) = sin 60

sin(180 – theta) = sin theta 

https://www.math-only-math.com/trigonometrical-ratios-of-180-degree-minus-theta.html

Or else use the calci to check it if u can’t remember it.

Answer 3

Here's a very simple way to solve this:

In Δ PQR --> Area(Δ PQR) = Area (Δ QPS) + Area( Δ PSR)

$\frac{1}{2}  r  q  \sin 120 = \frac{1}{2}  r  * PS  \sin 60 + \frac{1}{2}  q * PS \sin 60$

$\frac{1}{2}  r q * \sin (180-60) = \frac{1}{2}  PS \sin 60 * (r+q)$

$\frac{1}{2}  r q * \sin (60) = \frac{1}{2}  PS \sin 60 * (r+q)$

 

--> $PS = \frac{r * q}{r+q}$

 

Comments

See the LHS. It's $\frac{1}{2} rq sin 120$

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okay I got it I didn't know we can write area of triangle as half of product of any two sides * sin of angle between them thanks.

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mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html

Answer 4

Simple Method

So we get PS = r/2 ( or = q/2 ) by simple trigonometry

bcoz of angle property r = q 

now just put q = r in options 

A will give 2/r

B will give r/2 (That’s correct)

C will give r$\sqrt{2}$

D will give 4

Therefore answer is B

Answer 5

area of a triangle  =  1/2 ab sin theta , where a and b are two sides of a triangle and theta is the angle btw sides a and b

as, PS is the angle bisector of ∠QPR, so QPS = SPR = 60 or QPR = 120

similarly,

area of PQR = ar of PQS + ar of PSR

1/2 rq sin 120 = 1/2 * r * ps sin 60 + 1/2 * q * ps sin 60

option b (ans)