# Ex.6.2 Q4 Lines and Angles Solution - NCERT Maths Class 9

## Question

In the given figure, if \(PQ || ST\), \(\angle PQR = 110^ \circ\) and \(\angle RST = 130^ {\circ}\), find \(\angle QRS\).

## Text Solution

**What is known?**

\(PQ \| ST, \angle PQR = 110 ^ { \circ }\) and \( \angle RST = 130 ^ { \circ }\)

**What is unknown?**

\(\angle QRS =?\)

**Reasoning:**

- Lines which are parallel to the same line are parallel to each other.
- When two parallel lines are cut a transversal, co-interior angles formed are supplementary.

**Steps:**

Draw a line \(AB\) parallel to \(ST\) through point \(R\). Since \(AB\|ST\) and \(PQ\|ST\). So, \(AB\|PQ\).

Let \(\angle SRQ = x,\;\; \angle SRB = y\) and \(\angle QRA = z\)

Lines \(ST\) and \(AB\) are parallel with transversal \(SR\) intersecting. Therefore, the co-interior angles are supplementary.

\[\begin{align} \angle RST + \angle SRB & = 180 ^ { \circ } \\ 130 ^ { \circ } + y & = 180 ^ { \circ } \\ y & = (180 ^ { \circ } - 130 ^ { \circ }) \\ & = 50 ^ { \circ } \\ \angle SRB = y & = 50 ^ { \circ } \end{align}\]

Similarly, lines \(PQ\) and \(AB\) are parallel with transversal \(QR\) intersecting. Therefore, the co-interior angles are supplementary.

\[\begin{align} \angle PQR + \angle QRA & = 180 ^ { \circ } \\ 110 ^ { \circ } + z & = 180 ^ { \circ } \\ z & = (180 ^ { \circ } - 110 ^ { \circ }) \\ & = 70 ^ { \circ } \\ \angle QRA = z & = 70 ^ { \circ } \end{align}\]

\(AB\) is a line, \(RQ\) and \(RS\) are rays on \(AB\). Hence,

\[\begin{align} \angle QRA + \angle QRS + &\angle SRB = 180 ^ { \circ } \\ 70 ^ { \circ } + x + 50 ^ { \circ } & = 180 ^ { \circ } \\ 120 ^ { \circ } + x & = 180 ^ { \circ } \\ x & = (180 ^ { \circ } - 120 ^ { \circ }) \\ x & = 60 ^ { \circ } \\ \angle QRS = x & = 60 ^ { \circ } \end{align}\]