# Lines and Angles - NCERT Class 9 Maths

Exercise 6.1

## Question 1

In the given figure lines \(AB\) and \(CD\) intersect at \(O.\) If \(\begin{align} \angle AOC + \angle BOE = 70 ^ { \circ } \end{align}\) and \(\begin{align} \angle BOD = 40 ^ { \circ }, \end{align}\) find \(\angle BOE\) and reflex \(\angle COE\).

### Solution

**Video Solution**

**What is known?**

\(\angle AOC+\angle BOE={70}^{\circ}\) and \(\angle BOD = 40^{\circ}\)

**What is unknown?**

\(\angle BOE =?,\) Reflex \(\angle COE =?\)

**Reasoning:**

We know that vertically opposite angles formed when two lines intersect are equal. Also, sum of the adjacent angles is \(180\) degrees.

**Steps:**

Let \(\Delta AOC = \rm x\) and \(\rm{}\Delta BOE =\rm y.\)

Then \(x+ y=70^\circ\) \((\because AOC+BOE=70^\circ )\)

Let Reflex \(\Delta COE = \rm z\)

We can see that \(AB\) and \(CD\) are two intersecting lines so the pair of angles formed are vertically opposite angles and they are equal.

\(\text{i.e. } \angle AOD\) and \(\angle BOC\) and \(\angle AOC = \angle BOD.\)

Since \(\angle AOC = x\) and \(\angle AOC = \angle BOD = 40^ {\circ},\)

we can say that \(x = 40^ {\circ}.\)

Also we know that,

\[\begin{align} x + y &= 70 ^ {\circ } \\ 40 ^ { \circ } + y &= 70 ^ {\circ } \\ y &= 70 ^ { \circ } - 40 ^ {\circ } \\ y &= 30 ^ { \circ } \\ \angle BOE &= 30 ^ { \circ } \end{align}\]

If we consider line \(AB\) and ray \(OD\) on it, then \(\angle AOD\) and \(\angle BOD\) are adjacent angles.

\[\begin{align} \angle AOD + \angle BOD &= 180 ^ { \circ } \\ \angle AOD + 40 ^ { \circ } &= 180 ^ { \circ } \\ \angle AOD &= 180 ^ { \circ } - 40 ^ { \circ } \\ &= 140 ^ { \circ } \end{align}\]

\(\begin{align}\text{Reflex }\angle COE &=\begin{pmatrix}\angle AOC +\\ \angle AOD +\\ \angle BOD +\\ \angle BOE \end{pmatrix} \\ &= \begin{pmatrix}40 ^ { \circ } + 140 ^ { \circ } \\+ 40 ^ { \circ } + 30 ^ { \circ } \end{pmatrix}\\ &= 250 ^ { \circ } \end{align}\)

## Question 2

In the given figure, lines \(XY\) and \(MN\) intersect at \(O.\) If \(\angle POY = 90^ {\circ}\) and \(a:b = 2:3,\) find \(c.\)

### Solution

**Video Solution**

**What is known?**

\(\angle POY= 90^ {\circ}\) and \(a:b = 2:3\)

**What is unknown?**

\(\angle XON = c =?\)

**Reasoning:**

If two lines intersect each other, then the vertically opposite angles formed are equal.

**Steps:**

Line \(OP\) is perpendicular to line \(XY.\) Hence \(\angle POY = \angle POX = 90^ {\circ}\)

\(\begin{align}\angle POX & = \angle POM + \angle MOX\end{align}\)

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

Since \(a\) and \(b\) are in the ratio \(2:3\) that is, \(a = 2x\) and \(b = 3x \quad \ldots \ldots . (2)\)

Substituting (\(2\)) in (\(1\)),

\[\begin{align} a + b &= 90 ^ { \circ } \\ 2 x + 3 x &= 90 ^ { \circ } \\ 5 x &= 90 ^ { \circ } \\ x = \frac { 90 } { 5 } &= 18 ^ { \circ } \\ a = 2 x &= 2 \times 18 ^ { \circ } \\ a &= 36 ^ { \circ } \\ b = 3 x &= 3 \times 18 ^ { \circ } \\ b &= 54 ^ { \circ } \end{align}\]

Also,\(\begin{align}\angle MOY & = \angle MOP+ \angle POY\end{align}\)

\[\begin{align} & = a + 90 ^ { \circ } \\ & = 36 ^ { \circ } + 90 ^ { \circ } \\&= 126 ^ { \circ } \end{align}\]

Lines \(MN\) and \(XY\) intersect at point \(O\) and the vertically opposite angles formed are equal.

\[\begin{align}\angle XON &= \angle MOY \\ c &=126^ {\circ} \end{align} \]

## Question 3

In the given figure, \(\angle PQR = \angle PRQ\) then prove that \(\angle PQS = \angle PRT.\)

### Solution

**Video Solution**

**What is known?**

\(\angle PQR = \angle PRQ\)

**What is unknown?**

To prove \(\angle PQS = \angle PRT\)

**Reasoning:**

If a ray stands on a line, then the sum of adjacent angles formed is \(180^ {\circ}.\)

**steps:**

Let \(\angle PQR= \angle PRQ = a\)

Let \(\angle PQS = b \) and \(\angle PRT = c \)

Line \(ST \) and \(PQ\) intersect at point \(Q\), then the sum of adjacent angles \(\angle PQS\) and \(\angle PQR\) is \(180^ {\circ}.\)

\(\begin{align} \angle PQS + \angle PQR & = 180 ^ { \circ } \\ b + a & = 180 ^ { \circ } \\ b & = 180 ^ { \circ } - a\ldots . ( 1 ) \end{align}\)

Line \(ST\) and \(PR\) intersect at point \(R\), then the sum of adjacent angles \(\angle PRQ\) and \(\angle PRT \) is \(180^ {\circ}.\)

\(\begin{align} \angle PRQ + \angle PRT & = 180 ^ { \circ } \\ a + c & = 180 ^ { \circ } \\ c & = 180 ^ { \circ } - a\ldots . ( 2 ) \end{align}\)

From equations (\(1\)) and (\(2\)), it is clear that \(b = c.\)

Hence \(\angle PQS = \angle PRT\) is proved.

## Question 4

In the given figure, if \(x+y = w+z,\) then prove that \(AOB\) is a line.

### Solution

**Video Solution**

**What is known?**

\(\begin{align}x+y = w+z\end{align}\)

**What is unknown?**

To prove that \(AOB\) is a line.

**Reasoning:**

If the sum of two adjacent angles is \(180^ {\circ}\), then the non–common arms of the angles form a line.

**Steps:**

From the figure we can see that:

\(\begin{align}(x + y) +( w + z) = 360^ {\circ} \\ (\text {complete angle})\end{align}\)

It is given that \((x + y) = (w + z),\) Hence

\((x + y) + (w + z) = 360^ {\circ}\) can be written as \((x + y) + (x +y) = 360^ {\circ}\)

\[\begin{align} 2 x + 2 y &= 360 ^ { \circ } \\ 2 ( x + y ) &= 360 ^ { \circ } \\ x + y &= \frac { 360 } { 2 } \\&= 180 ^ { \circ } \end{align}\]

Since sum of adjacent angles \(x\) and \(y\) with \(OA\) and \(OB\) as the non- common arms is \(180^ {\circ}\) we can say that \(AOB\) is a line.

## Question 5

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 \(\begin{align} \angle ROS =\frac {1}{2}(\angle QOS - \angle POS).\end{align}\)

### Solution

**Video 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 (\(2\)),

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

## Question 6

It is given that \(\angle XYZ = 64^ {\circ}\) and \(XY\) is produced to point \(P.\) Draw a figure from the given information. If ray \(YQ\) bisects \(\angle ZYP,\) find \(\angle XYQ\) and reflex \(\angle QYP.\)

### Solution

**Video Solution**

**What is known?**

\(\angle XYZ = 64^ {\circ} \) and Ray \(YQ\) bisects \(\angle PYZ.\)

**What is unknown?**

\(\angle XYQ = ?\) and Reflex \(\angle QYP = ?\)

**Reasoning:**

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

**Steps:**

With the given information in the question, we can come up with this diagram.

Ray \(YQ\) bisects \(\angle ZYP\) and let \(\angle ZYQ = \angle QYP = a.\)

We can see from figure that \(PX\) a line and \(YZ\) is a ray intersecting at point \(Y\) and the sum of adjacent angles so formed is \(180^ {\circ}.\)

Hence \( \angle ZYP + \angle ZYX = 180 ^ { \circ }\)

\(\begin{align}\angle ZYQ + \angle QYP + &\angle ZYX = 180 ^ { \circ } \\ a + a + 64 ^ {\circ } & = 180 ^ { \circ } \\ 2 a + 64 ^ {\circ } & = 180 ^ { \circ } \\ 2 a & = 180 ^ { \circ }-64 ^ {\circ } \\&= 116 ^ { \circ } \\ a & = \frac { 116 } { 2 } \\ & = 58 ^ { \circ } \end{align}\)

Then \(\angle XYQ = \angle XYZ + \angle ZYQ\)

\[\begin{align}\angle XYQ &= a + 64 ^ {\circ } \\ &= 58 ^ { \circ } + 64 ^ {\circ} \\ &= 122 ^ { 0 } \end{align}\]

As,

\(\begin{align}\angle QYP &= a, \text {Reflex}\angle QYP \\&= (360 ^ { \circ } - a) \end{align}\)

\[\begin{align}&= (360 ^ { \circ } - 58 ^ { \circ }) \\ &= 302 ^ { \circ } \end{align}\]

Reflex,\(\angle QYP = 302 ^ { \circ }\)

The chapter 6 starts with an introduction about points and lines we covered in previous grades followed by basic terms and definitions used. Next, we have a brief introduction of intersecting lines and non-intersecting lines with an explanation to a pair of angles. Later, the chapter discusses parallel lines and transversal with various theorems involved. Finally, in the end, the chapter focuses on the angle sum property of a triangle with few examples and a series of exercise questions.

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