Lines and Angles
NCERT Solutions • Class 9 Maths • Chapter 6Exercise 6.1
1. In Fig. 6.13, lines AB and CD intersect at O. If $\angle AOC + \angle BOE = 70^\circ$ and $\angle BOD = 40^\circ$, find $\angle BOE$ and reflex $\angle COE$.
Step 1: Lines AB and CD intersect at O.
$\angle AOC = \angle BOD$ (Vertically opposite angles)
Given $\angle BOD = 40^\circ$, therefore $\angle AOC = 40^\circ$.
$\angle AOC = \angle BOD$ (Vertically opposite angles)
Given $\angle BOD = 40^\circ$, therefore $\angle AOC = 40^\circ$.
Step 2: Find $\angle BOE$.
Given $\angle AOC + \angle BOE = 70^\circ$.
$40^\circ + \angle BOE = 70^\circ \Rightarrow \angle BOE = 70^\circ – 40^\circ$
$\angle BOE = 30^\circ$
Given $\angle AOC + \angle BOE = 70^\circ$.
$40^\circ + \angle BOE = 70^\circ \Rightarrow \angle BOE = 70^\circ – 40^\circ$
$\angle BOE = 30^\circ$
Step 3: Find Reflex $\angle COE$.
Since AOB is a straight line: $\angle AOC + \angle COE + \angle BOE = 180^\circ$
$40^\circ + \angle COE + 30^\circ = 180^\circ \Rightarrow \angle COE = 180^\circ – 70^\circ = 110^\circ$.
Reflex $\angle COE = 360^\circ – 110^\circ =$ $250^\circ$.
Since AOB is a straight line: $\angle AOC + \angle COE + \angle BOE = 180^\circ$
$40^\circ + \angle COE + 30^\circ = 180^\circ \Rightarrow \angle COE = 180^\circ – 70^\circ = 110^\circ$.
Reflex $\angle COE = 360^\circ – 110^\circ =$ $250^\circ$.
2. In Fig. 6.14, lines XY and MN intersect at O. If $\angle POY = 90^\circ$ and $a : b = 2 : 3$, find $c$.
Step 1: Calculate $a$ and $b$.
Since $\angle POY = 90^\circ$, angles on line XY sum to $180^\circ$:
$\angle POX = 180^\circ – 90^\circ = 90^\circ$.
Also, $\angle POX = a + b$. So, $a + b = 90^\circ$.
Given ratio $a:b = 2:3$. Let $a = 2x, b = 3x$.
$2x + 3x = 90^\circ \Rightarrow 5x = 90^\circ \Rightarrow x = 18^\circ$.
$a = 2(18) = 36^\circ$, $b = 3(18) = 54^\circ$.
Since $\angle POY = 90^\circ$, angles on line XY sum to $180^\circ$:
$\angle POX = 180^\circ – 90^\circ = 90^\circ$.
Also, $\angle POX = a + b$. So, $a + b = 90^\circ$.
Given ratio $a:b = 2:3$. Let $a = 2x, b = 3x$.
$2x + 3x = 90^\circ \Rightarrow 5x = 90^\circ \Rightarrow x = 18^\circ$.
$a = 2(18) = 36^\circ$, $b = 3(18) = 54^\circ$.
Step 2: Find $c$.
Line MN is a straight line. Therefore, $b + c = 180^\circ$ (Linear Pair).
$54^\circ + c = 180^\circ \Rightarrow c = 180^\circ – 54^\circ$
$c = 126^\circ$.
Line MN is a straight line. Therefore, $b + c = 180^\circ$ (Linear Pair).
$54^\circ + c = 180^\circ \Rightarrow c = 180^\circ – 54^\circ$
$c = 126^\circ$.
3. In Fig. 6.15, $\angle PQR = \angle PRQ$, then prove that $\angle PQS = \angle PRT$.
Proof:
1. $\angle PQS + \angle PQR = 180^\circ$ (Linear Pair) $\Rightarrow \angle PQR = 180^\circ – \angle PQS$ … (i)
2. $\angle PRT + \angle PRQ = 180^\circ$ (Linear Pair) $\Rightarrow \angle PRQ = 180^\circ – \angle PRT$ … (ii)
3. Given $\angle PQR = \angle PRQ$.
Substituting (i) and (ii): $180^\circ – \angle PQS = 180^\circ – \angle PRT$
$- \angle PQS = – \angle PRT \Rightarrow$ $\angle PQS = \angle PRT$. (Hence Proved)
1. $\angle PQS + \angle PQR = 180^\circ$ (Linear Pair) $\Rightarrow \angle PQR = 180^\circ – \angle PQS$ … (i)
2. $\angle PRT + \angle PRQ = 180^\circ$ (Linear Pair) $\Rightarrow \angle PRQ = 180^\circ – \angle PRT$ … (ii)
3. Given $\angle PQR = \angle PRQ$.
Substituting (i) and (ii): $180^\circ – \angle PQS = 180^\circ – \angle PRT$
$- \angle PQS = – \angle PRT \Rightarrow$ $\angle PQS = \angle PRT$. (Hence Proved)
4. In Fig. 6.16, if $x + y = w + z$, then prove that AOB is a line.
Proof:
Sum of all angles around a point is $360^\circ$.
$x + y + w + z = 360^\circ$
Given $x + y = w + z$. Substitute $(w+z)$ with $(x+y)$:
$(x + y) + (x + y) = 360^\circ \Rightarrow 2(x + y) = 360^\circ \Rightarrow x + y = 180^\circ$.
Since angles $x$ and $y$ form a linear pair (sum is $180^\circ$), AOB is a straight line.
Sum of all angles around a point is $360^\circ$.
$x + y + w + z = 360^\circ$
Given $x + y = w + z$. Substitute $(w+z)$ with $(x+y)$:
$(x + y) + (x + y) = 360^\circ \Rightarrow 2(x + y) = 360^\circ \Rightarrow x + y = 180^\circ$.
Since angles $x$ and $y$ form a linear pair (sum is $180^\circ$), AOB is a straight line.
5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\angle ROS = \frac{1}{2}(\angle QOS – \angle POS)$.
Proof:
Since $OR \perp PQ$, $\angle POR = \angle QOR = 90^\circ$.
1. $\angle ROS = \angle POR – \angle POS = 90^\circ – \angle POS$ … (i)
2. $\angle ROS = \angle QOS – \angle QOR = \angle QOS – 90^\circ$ … (ii)
Adding (i) and (ii):
$2\angle ROS = (90^\circ – \angle POS) + (\angle QOS – 90^\circ)$
$2\angle ROS = \angle QOS – \angle POS$
$\angle ROS = \frac{1}{2}(\angle QOS – \angle POS)$ (Hence Proved)
Since $OR \perp PQ$, $\angle POR = \angle QOR = 90^\circ$.
1. $\angle ROS = \angle POR – \angle POS = 90^\circ – \angle POS$ … (i)
2. $\angle ROS = \angle QOS – \angle QOR = \angle QOS – 90^\circ$ … (ii)
Adding (i) and (ii):
$2\angle ROS = (90^\circ – \angle POS) + (\angle QOS – 90^\circ)$
$2\angle ROS = \angle QOS – \angle POS$
$\angle ROS = \frac{1}{2}(\angle QOS – \angle POS)$ (Hence Proved)
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 $\angle QYP$.
Step 1: Find $\angle ZYP$.
Since XY is produced to P, XYP is a line.
$\angle XYZ + \angle ZYP = 180^\circ$ (Linear Pair)
$64^\circ + \angle ZYP = 180^\circ \Rightarrow \angle ZYP = 116^\circ$.
Since XY is produced to P, XYP is a line.
$\angle XYZ + \angle ZYP = 180^\circ$ (Linear Pair)
$64^\circ + \angle ZYP = 180^\circ \Rightarrow \angle ZYP = 116^\circ$.
Step 2: Find $\angle XYQ$.
YQ bisects $\angle ZYP$, so $\angle ZYQ = \angle QYP = \frac{116^\circ}{2} = 58^\circ$.
$\angle XYQ = \angle XYZ + \angle ZYQ = 64^\circ + 58^\circ =$ $122^\circ$.
YQ bisects $\angle ZYP$, so $\angle ZYQ = \angle QYP = \frac{116^\circ}{2} = 58^\circ$.
$\angle XYQ = \angle XYZ + \angle ZYQ = 64^\circ + 58^\circ =$ $122^\circ$.
Step 3: Find Reflex $\angle QYP$.
Reflex $\angle QYP = 360^\circ – \angle QYP$
$= 360^\circ – 58^\circ =$ $302^\circ$.
Reflex $\angle QYP = 360^\circ – \angle QYP$
$= 360^\circ – 58^\circ =$ $302^\circ$.