# Ex.6.3 Q5 Lines and Angles Solution - NCERT Maths Class 9

## Question

In Fig. below, if \(PQ ⊥ PS\) , \(PQ || SR\) ,\(\angle SQR = 28^\circ \) and \(\angle QRT = 65^\circ \) ,then find the values of *\(x\)* and *\(y\)* .

## Text Solution

**What is known?**

\(PQ ⊥ PS,\, PQ || SR\), \(\angle SQR = 28^\circ \) and \(\angle QRT = 65^\circ \)

**What is unknown?**

*\(x\)* and *\(y\)*.

**Reasoning:**

As we know when two parallel lines are cut by a transversal, alternate interior angles formed are equal.

Angle sum property of a triangle:

Sum of the interior angles of a triangle is \(360^\circ\).

**Steps:**

Given,

\(PQ ⊥ PS,\, PQ || SR\), \(\angle SQR = 28^\circ \) and \(\angle QRT = 65^\circ \)

\(\begin{align}\angle PQR &= \angle QRT\\ \begin{array}{I}(\text{Alternate }\end{array} &\begin{array}{I}\text{interior angles}) \end{array}\\\\ \angle PQS + \angle SQR &= \angle QRT \\ \text{(By figure)}\\\\x + 28^\circ &= 65^\circ \\x &= 65^\circ - 28^\circ \\x &= 37^\circ \end{align}\)

Now, in \(\Delta PQS\)

\(\begin{align}\angle PQS + \angle PSQ + \angle QPS &=180^\circ\\ ( \text{Angle sum property of a }& \text{triangle}\rm{.})\\ \\37^\circ + y + 90^\circ &= 180^\circ \\y &=180^\circ \!- \!127^\circ \\y &=\, 53^\circ \end{align}\)

Hence, \(x = 37^\circ \) and \(y = 53^\circ \)