in Quantitative Aptitude edited by
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26 votes
26 votes

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

  1. $\left(\dfrac{(q+r)} {qr}\right)$
  2. $\left(\dfrac {qr} {q+r}\right)$
  3. $\large \sqrt {(q^2 + r^2)}$
  4. $\left(\dfrac{(q+r)^2} {qr}\right)$
in Quantitative Aptitude edited by
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4 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°
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3 Answers

39 votes
39 votes
Best answer

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

edited by
by

4 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.

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68 votes
68 votes

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..

edited by

2 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
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17 votes
17 votes

Here's a very simple way to solve this:

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

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

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

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

 

--> $PS = \frac{r * q}{r+q}$

 

edited by

4 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
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Answer:

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