# Ex.6.1 Q5 Lines and Angles Solution - NCERT Maths Class 9

## Question

In the given figure, \(POQ\) is a line. Ray \(OR\) is perpendicular to line \(PQ.\;OS\) another ray lying between rays \(OP\) and \(OR.\) Prove that \(\begin{align} \angle ROS =\frac {1}{2}(\angle QOS - \angle POS).\end{align}\)

## Text Solution

**What is known?**

\(OR\) is perpendicular to \(PQ.\) \(\angle ROQ = \angle ROP =90^ {\circ}.\)

**What is unknown?**

To prove that: \(\begin{align} \angle ROS =\frac {1}{2}(\angle QOS - \angle POS)\end{align}\)

**Reasoning:**

When a ray intersects a line, then the sum of adjacent angles so formed is \(180^ {\circ}.\)

**Steps:**

Let \(\angle ROS = a, \angle POS = b\) and \(\angle SOQ = c.\)

To prove that: \(\begin{align}a= \frac{1}{2}(c-b).\end{align}\)

Since \(\angle ROQ = \angle ROP = 90 ^ { \circ }\)

We can say,

\(\begin{align} \angle POS + \angle SOR &= \angle POR \end{align}\)

\[\begin{align}\rm{} b + a &= 90 ^ { \circ } \ldots \ldots \ldots ( 1 ) \end{align}\]

Line \(PQ\) is intersected by ray \(OS.\)

Hence, \(\begin{align}\angle POS + \angle SOQ = \,& b + c = 180 ^ { \circ } \end{align}\)

\[ \begin{align}& b + c = 180 ^ { \circ } \ldots \ldots \ldots (2) \end{align}\]

From equation (1), we get: \(a + b = 90^ {\circ}\)

Multiplying by \(2\) on both sides we get,

\[\begin{align} 2 ( a + b ) &= 2 \times 90 ^ { \circ } \\ 2 ( a + b ) &= 180 ^ { \circ } \ldots \ldots \ldots ( 3 ) \end{align}\]

Comparing equations (\(3\)) and (\(2\)),

\[\begin{align} 2 ( a + b ) &= b + c \\ 2 a + 2 b &= b + c \\ 2 a &= b + c - 2 b \\ 2 a &= c - b \\ a &= \frac { 1 } { 2 } ( c - b ) \\ \therefore \angle ROS &= \frac { 1 } { 2 } ( \angle QOS - \angle POS). \end{align}\]