NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

Question 1: In Fig. 6.13, lines AB and CD intersect at O. If $\angle$ AOC + $\angle$ BOE = 70° and $\angle$ BOD = 40°, find $\angle$ BOE and reflex $\angle$ COE.

Answer:

Given that,

AB is a straight line. Lines AB and CD intersect at O.

$\angle AOC + \angle BOE = 70°$ and $\angle$ BOD = $40°$

Since AB is a straight line,

$\therefore$ $\angle$ AOC + $\angle$ COE + $\angle$ EOB = $180°$

$\Rightarrow \angle COE = 180°-70°=110°$

[since $\angle AOC + \angle BOE = 70°$ ]

So, reflex $\angle$ COE = $360°-110° = 250°$

It is given that AB and CD intersect at O.

Therefore, $\angle$ AOC = $\angle$ BOD [vertically opposite angle]

$\Rightarrow \angle COA = 40°$ [ GIven $\angle$ BOD = $40°$ ]

Also, $\angle AOC + \angle BOE = 70°$

So, $\angle$ BOE = $30°$

Question 3: In Fig. 6.15, $\angle$ PQR = $\angle$ PRQ, then prove that $\angle$ PQS = $\angle$ PRT.

Answer:

Given that,

ABC is a triangle such that $\angle$ PQR = $\angle$ PRQ, and ST is a straight line.

Now, $\angle$ PQR + $\angle$ PQS = $180°$ {Linear pair}............(I)

Similarly, $\angle$ PRQ + $\angle$ PRT = $180°$ ..................(ii)

Equating the equation (i) and eq (ii), we get,

$\angle PQR +\angle PQS =\angle PRT + \angle PRQ$ {but $\angle$ PQR = $\angle$ PRQ }

Therefore, $\angle$ PQS = $\angle$ PRT

Hence, it is proved.

Question 4: In Fig. 6.16, ifx + y = w + z, then prove that AOB is a line.

Answer:

Given that,

$x+y = z+w$ ..............(I)

It is known that the sum of all the angles at a point = $360°$

$\therefore$ $x+y+z+w=360°$ ..............(ii)

From eq (i) and eq (ii), we get,

$\\2(x+y)=360°$

$\Rightarrow x+y = 180°$

Hence, proven that AOB is a line.

Question 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 $\angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})$

Answer:

Given that,

POQ is a line, OR $\perp$ PQ and $\angle$ ROQ is a right angle.

Now, $\angle$ POS + $\angle$ ROS + $\angle$ ROQ = $180°$ [since POQ is a straight line]

$\\\Rightarrow \angle POS + \angle ROS = 90°$

$ \Rightarrow \angle ROS = 90°-\angle POS$ .............(I)

and, $\angle$ ROS + $\angle$ ROQ = $\angle$ QOS

$\angle ROS = \angle QOS -90°$ ..............(ii)

Adding the eq (i ) and eq (ii), we get,

$\angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})$

Hence, proved.

Question 6: It is given that $\angle$ XYZ = 64° 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.

Answer:

Given that,

$\angle$ XYZ = $64°$ and XY produced to point P and Ray YQ bisects $\angle$ ZYP $\Rightarrow \angle QYP = \angle ZYQ$

Now, XYP is a straight line.

So, $\angle$ XYZ + $\angle$ ZYQ + $\angle$ QYP = $180°$

$\Rightarrow 2(\angle QYP )=180° - 64° = 116°$

$\Rightarrow (\angle QYP )= 58°$

Thus reflex of $\angle$ QYP = $360°- 58° = 302°$

Since $\angle$ XYQ = $\angle$ XYZ + $\angle$ ZYQ

[ $\because$ $\angle QYP = \angle ZYQ$]

$\angle XYQ$ = $64°+58° = 122°$

Question 1: In Fig. 6.29, if AB $||$ CD, CD $||$ EF and y : z = 3 : 7 , find x .

Answer:

Given AB || CD and CD || EF and $y:z = 3:7$

$ \Rightarrow z = \frac{7}{3}y$

Therefore, AB || EF and $x =z$ (alternate interior angles)..............(i)

Again, CD || AB

$\\\Rightarrow x+y =180°$

$ \Rightarrow z+y =180°$ .............(ii)

Putting the value of $z$ in equation (ii), we get,

$\frac{10}{3}y =180°$

$ \Rightarrow y =54°$

Then $z=180°-54°=126°$

By equation (i), we get the value of $x=126°$

Question 2: In Fig. 6.30, if AB $||$ CD, EF $\bot$ CD and $\angle$ GED = 126°, find $\angle$ AGE, $\angle$ GEF and $\angle$ FGE.

Answer:

Given AB || CD, EF $\perp$ CD and $\angle$ GED = $126°$

In the above figure,

GE is transversal.

So, that $\angle$ AGE = $\angle$ GED = $126°$

[Alternate interior angles]

Also, $\angle$ GEF = $\angle$ GED - $\angle$ FED = $126°-90° = 36°$

$\Rightarrow \angle GEF = 36°$

Since AB is a straight line.

Therefore, $\angle$ AGE + $\angle$ FGE = $180°$

So, $\angle$ FGE = $180°-\angle AGE = 180° - 126°$

$\Rightarrow \angle FGE =54°$

Question 3: In Fig. 6.31, if PQ $||$ ST, $\angle$ PQR = 110° and $\angle$ RST = 130°, find $\angle$ QRS. [Hint: Draw a line parallel to ST through point R.]

Answer:

Draw a line EF parallel to the ST through R.

Since PQ || ST and ST || EF

$\Rightarrow$ EF || PQ

$\angle$ PQR = $\angle$ QRF = $110°$(Alternate interior angles)

$\angle$ QRF = $\angle$ QRS + $\angle$ SRF .............(i)

Again, $\angle$ RST + $\angle$ SRF = $180°$ (Interior angles of two parallels ST and RF)

$\Rightarrow \angle SRF =180°-130° = 50°$ ( $\angle$ RST = $130°$ , given)

Thus, $\angle$ QRS = $110°-50° = 60°$

Question 4: In Fig. 6.32, if AB $||$ CD, $\angle$ APQ = 50° and $\angle$ PRD = 127°, find x and y.

Answer:

Given, AB || CD, $\angle$ APQ = $50°$ and $\angle$ PRD = $127°$

PQ is a transversal.

So,

$\angle$ APQ = $\angle$ PQR= $50°$ (alternate interior angles)

$\therefore x = 50°$

Again, PR is a transversal.

So,

$\angle$ y + $50°$ = $127°$ (Alternate interior angles)

$\Rightarrow y = 77°$

Question 5: In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B; the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects along CD. Prove that AB $||$ CD.

Answer:

Draw a ray BL $\perp$ PQ and CM $\perp$ RS

Since PQ || RS (Given)

So, BL || CM and BC are a transversal.

$\therefore$ $\angle$ LBC = $\angle$ MCB (Alternate interior angles).............(I)

It is known that angle of incidence = angle of reflection

So, $\angle$ ABL = $\angle$ LBC and $\angle$ MCB = $\angle$ MCD

$\Rightarrow \angle ABL =\angle MCD$ ..................(ii)

Adding equations (i) and (ii), we get,

$\angle$ ABC = $\angle$ DCB

Both the interior angles are equal

Hence AB || CD