NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

Q1. In Fig. 6.13, lines AB and CD intersect at O. If $\mathrm{\angle }$ AOC + $\mathrm{\angle }$ BOE = 70° and $\mathrm{\angle }$ BOD = 40°, find $\mathrm{\angle }$ BOE and reflex $\mathrm{\angle }$ COE.

Answer:

Given that,
AB is a straight line. Lines AB and CD intersect at O.
$\mathrm{\angle }AOC+\mathrm{\angle }BOE=70°$ and $\mathrm{\angle }$ BOD = $40°$
Since AB is a straight line,
$\therefore$ $\mathrm{\angle }$ AOC + $\mathrm{\angle }$ COE + $\mathrm{\angle }$ EOB = $180°$
$\Rightarrow \mathrm{\angle }COE=180°-70°={110}^0$
[since $\mathrm{\angle }AOC+\mathrm{\angle }BOE=70°$ ]

So, reflex $\mathrm{\angle }$ COE = $360°-{110}^0=250°$
It is given that AB and CD intersect at O.
Therefore, $\mathrm{\angle }$ AOC = $\mathrm{\angle }$ BOD [vertically opposite angle]
$\Rightarrow \mathrm{\angle }COA=40°$ [ GIven $\mathrm{\angle }$ BOD = ${40}^0$ ]
Also, $\mathrm{\angle }AOC+\mathrm{\angle }BOE=70°$
So, $\mathrm{\angle }$ BOE = $30°$

Q2. In Fig. 6.14, lines XY and MN intersect at O. If $\mathrm{\angle }POY={90}^o$ and a: b = 2 : 3, find c.

Answer:

Given that,
Line XY and MN intersect at O and $\mathrm{\angle }$ POY = ${90}^0$
Also, $a:b=2:3\Rightarrow b=\frac{3a}{2}$ ..............(i)

Since XY is a straight line
Therefore, $a+b+\mathrm{\angle }POY={180}^0$
$\Rightarrow a+b={180}^0-{90}^0={90}^0$ ...........(ii)
Thus, from eq (i) and eq (ii), we get,
$\Rightarrow \frac{3a}{2}+a={90}^0$
$\Rightarrow a={36}^0$
So, $b={54}^0$
Since $\mathrm{\angle }$ MOY = $\mathrm{\angle }$ c [vertically opposite angles]
$\mathrm{\angle }$ a + $\mathrm{\angle }$ POY = c
${126}^0=c$

Q3. In Fig. 6.15, $\mathrm{\angle }$ PQR = $\mathrm{\angle }$ PRQ, then prove that $\mathrm{\angle }$ PQS = $\mathrm{\angle }$ PRT.

Answer:

Given that,
ABC is a triangle such that $\mathrm{\angle }$ PQR = $\mathrm{\angle }$ PRQ, and ST is a straight line.
Now, $\mathrm{\angle }$ PQR + $\mathrm{\angle }$ PQS = ${180}^0$ {Linear pair}............(i)
Similarly, $\mathrm{\angle }$ PRQ + $\mathrm{\angle }$ PRT = ${180}^0$ ..................(ii)

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

$\mathrm{\angle }PQR+\mathrm{\angle }PQS=\mathrm{\angle }PRT+\mathrm{\angle }PRQ$ {but $\mathrm{\angle }$ PQR = $\mathrm{\angle }$ PRQ }
Therefore, $\mathrm{\angle }$ PQS = $\mathrm{\angle }$ PRT
Hence, it is proved.

Q4. In Fig. 6.16, if x + 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 = ${360}^0$
$\therefore$ $x+y+z+w={360}^0$ ..............(ii)

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

$2(x+y)={360}^0$
$\Rightarrow x+y={180}^0$

Hence, proven that AOB is a line.

Q5. 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 $\mathrm{\angle }\text{ROS}=\frac{1}{2}(\mathrm{\angle }\text{QOS}-\mathrm{\angle }\text{POS})$

Answer:

Given that,
POQ is a line, OR $\perp$ PQ and $\mathrm{\angle }$ ROQ is a right angle.
Now, $\mathrm{\angle }$ POS + $\mathrm{\angle }$ ROS + $\mathrm{\angle }$ ROQ = ${180}^0$ [since POQ is a straight line]
$\Rightarrow \mathrm{\angle }POS+\mathrm{\angle }ROS={90}^0$
$\Rightarrow \mathrm{\angle }ROS={90}^0-\mathrm{\angle }POS$ .............(i)
and, $\mathrm{\angle }$ ROS + $\mathrm{\angle }$ ROQ = $\mathrm{\angle }$ QOS
$\mathrm{\angle }ROS=\mathrm{\angle }QOS-{90}^0$ ..............(ii)

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

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

Hence, proved.

Q6. It is given that $\mathrm{\angle }$ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects $\mathrm{\angle }$ ZYP, find $\mathrm{\angle }$ XYQ and reflex $\mathrm{\angle }$ QYP.

Answer:

Given that,
$\mathrm{\angle }$ XYZ = ${64}^0$ and XY produced to point P and Ray YQ bisects $\mathrm{\angle }$ ZYP $\Rightarrow \mathrm{\angle }QYP=\mathrm{\angle }ZYQ$
Now, XYP is a straight line.
So, $\mathrm{\angle }$ XYZ + $\mathrm{\angle }$ ZYQ + $\mathrm{\angle }$ QYP = ${180}^0$
$\Rightarrow 2(\mathrm{\angle }QYP)={180}^0-{64}^0={116}^0$
$\Rightarrow (\mathrm{\angle }QYP)={58}^0$

Thus reflex of $\mathrm{\angle }$ QYP = ${360}^0-{58}^0={302}^0$

Since $\mathrm{\angle }$ XYQ = $\mathrm{\angle }$ XYZ + $\mathrm{\angle }$ ZYQ
[ $\because$ $\mathrm{\angle }QYP=\mathrm{\angle }ZYQ$

$\mathrm{\angle }XYQ$ = ${64}^0+{58}^0={122}^0$

Q1. 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
$\Rightarrow x+y={180}^0$
$\Rightarrow z+y={180}^0$ .............(ii)

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

$\frac{10}{3}y={180}^0$
$\Rightarrow y={54}^0$
Then $z={180}^0-{54}^0={126}^0$

By equation (i), we get the value of $x={126}^0$

Q2. In Fig. 6.30, if AB $||$ CD, EF $\mathrm{\perp }$ CD and $\mathrm{\angle }$ GED = 126°, find $\mathrm{\angle }$ AGE, $\mathrm{\angle }$ GEF and $\mathrm{\angle }$ FGE.

Answer:

Given AB || CD, EF $\perp$ CD and $\mathrm{\angle }$ GED = ${126}^0$
In the above figure,
GE is transversal.
So, that $\mathrm{\angle }$ AGE = $\mathrm{\angle }$ GED = ${126}^0$
[Alternate interior angles]
Also, $\mathrm{\angle }$ GEF = $\mathrm{\angle }$ GED - $\mathrm{\angle }$ FED = ${126}^0-{90}^0={36}^0$
$\Rightarrow \mathrm{\angle }GEF={36}^0$

Since AB is a straight line.
Therefore, $\mathrm{\angle }$ AGE + $\mathrm{\angle }$ FGE = ${180}^0$
So, $\mathrm{\angle }$ FGE = ${180}^0-\mathrm{\angle }AGE={180}^0-{126}^0$
$\Rightarrow \mathrm{\angle }FGE={54}^0$

Q3. In Fig. 6.31, if PQ $||$ ST, $\mathrm{\angle }$ PQR = 110° and $\mathrm{\angle }$ RST = 130°, find $\mathrm{\angle }$ 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

$\mathrm{\angle }$ PQR = $\mathrm{\angle }$ QRF = ${110}^0$(Alternate interior angles)
$\mathrm{\angle }$ QRF = $\mathrm{\angle }$ QRS + $\mathrm{\angle }$ SRF .............(i)

Again, $\mathrm{\angle }$ RST + $\mathrm{\angle }$ SRF = ${180}^0$ (Interior angles of two parallels ST and RF)
$\Rightarrow \mathrm{\angle }SRF={180}^0-{130}^0={50}^0$ ( $\mathrm{\angle }$ RST = ${130}^0$ , given)

Thus, $\mathrm{\angle }$ QRS = ${110}^0-{50}^0={60}^0$

Q4. In Fig. 6.32, if AB $||$ CD, $\mathrm{\angle }$ APQ = 50° and $\mathrm{\angle }$ PRD = 127°, find x and y.

Answer:

Given, AB || CD, $\mathrm{\angle }$ APQ = ${50}^0$ and $\mathrm{\angle }$ PRD = ${127}^0$
PQ is a transversal.
So,
$\mathrm{\angle }$ APQ = $\mathrm{\angle }$ PQR= ${50}^0$ (alternate interior angles)
$\therefore x={50}^0$

Again, PR is a transversal.
So,
$\mathrm{\angle }$ y + ${50}^0$ = ${127}^0$ (Alternate interior angles)
$\Rightarrow y={77}^0$

Q5. 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 is a transversal.
$\therefore$ $\mathrm{\angle }$ LBC = $\mathrm{\angle }$ MCB (Alternate interior angles).............(i)

It is known that angle of incidence = angle of reflection
So, $\mathrm{\angle }$ ABL = $\mathrm{\angle }$ LBC and $\mathrm{\angle }$ MCB = $\mathrm{\angle }$ MCD
$\Rightarrow \mathrm{\angle }ABL=\mathrm{\angle }MCD$ ..................(ii)

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

$\mathrm{\angle }$ ABC = $\mathrm{\angle }$ DCB
Both the interior angles are equal
Hence AB || CD