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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.**

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.

0

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**..

17 votes

Here's a very simple way to solve this:

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

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

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

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

--> $PS = \frac{r * q}{r+q}$

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