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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. $\left(\dfrac{(q+r)^2} {qr}\right)$
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4 Answers

Best answer
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81 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..

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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
19 votes
19 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
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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:

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