Question

In a $\Delta PQR$, PS is bisector $\angle P$, $\angle Q={{70}^{\circ }},\angle R={{30}^{\circ }}$, then
(a) QS > PQ > PR
(b) QS < PQ < PR
(c) PQ > QS > SR
(d) PQ < QS < SR

Answer 1

Hint: PS bisects $\angle P$ into two equal halves. Thus find $\angle P$. Hence compare angles PQS, QPS, PSQ and PRQ to find out the sides which are larger. The side opposite to the larger angle will be longer and vice-versa.

Complete step-by-step answer:

From the figure, you can find $\Delta PQR$. It is said PS is the bisector of $\angle P$, which means that the line or line segment PS divides the $\angle P$ into two equal halves.

Thus, from the figure, we can say that, \[\angle QPS=\angle RPS,\angle P=\angle QPS+\angle RPS\]

\[\Rightarrow \angle P=\angle QPS+\angle QPS\]

\[\Rightarrow \angle P=Q\angle QPS\]…………………. (i)

Now, we have been given two angles of \[\Delta PQR\] , which are \[\angle Q={{70}^{\circ }},\angle R={{30}^{\circ }}\].

Thus with the help of their angles we can find \[\angle P\].

By using the triangle angle sum theorem, we can say that the measure of the angles of a triangle is \[{{180}^{\circ }}\]. Thus from $\Delta PQR$,

$\angle P+\angle Q+\angle R={{180}^{\circ }}$

Now let us put values of $\angle Q={{70}^{\circ }},\angle R={{30}^{\circ }}$in the above expression.

$\begin{align}

  & \therefore \angle P+70+30={{180}^{\circ }} \\

 & \angle P=180-100 \\

 & ={{80}^{\circ }} \\

\end{align}$

Hence, we got $\angle P={{80}^{\circ }}$. Now put this in (i)

$\begin{align}

  & \angle P=2\angle QPS \\

 & \angle QPS=\angle RPS=\dfrac{\angle P}{2}=\dfrac{{{80}^{\circ }}}{2}={{40}^{\circ }} \\

 & \therefore \angle QPS=\angle RPS={{40}^{\circ }} \\

\end{align}$

By looking the figure, you can say that $\angle PRQ\to {{30}^{\circ }}$ is less than $\angle PQR\to {{70}^{\circ }}$. i.e., $\angle PRQ<\angle PQR$

Hence from this we can say that the length of side PQ is less than length of side PR.
PQ < PR ………………. (ii)

Now $\angle QPS<\angle QPS,\angle PQS={{70}^{\circ }}$

i.e. $\angle QPS<\angle PQS$

Thus, QS < PR ……………………. (iii)

Now, from the figure, let us consider $\Delta PQS$. By triangle angle sum theorem, we can say that

$\begin{align}

  & \angle QPS+\angle PQS+\angle PSQ={{180}^{\circ }} \\

 & {{70}^{\circ }}+{{40}^{\circ }}+\angle PSQ={{180}^{\circ }} \\

 & \angle PQS={{180}^{\circ }}-{{110}^{\circ }}={{70}^{\circ }} \\

\end{align}$

Thus we got $\angle PQS={{70}^{\circ }}$

Hence angle QPS = ${{40}^{\circ }}$ is less than $\angle PQS={{70}^{\circ }}$

i.e., $\angle QPS<\angle PSQ$

Thus we can say that QS < PS ……………….. (iv)

Now let us consider (ii), (iii), (iv)

PQ < PR, QS < PR, QS < PQ .

From the above we can say that PR is greater than both PQ,QS.

PQ is greater than QS. Hence, we can say that

QS < PQ < PR.

Hence we got the required answer.

Option (b) is the correct answer.

Note: If angle is a triangle is larger than another angle in a triangle, then the side opposite to the larger angle will be longer than the side opposite to the smaller angle, this is the concept we have used here. Hence, if angle is given, compare them to find the length which is larger.