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#### Question 31:

In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

$\angle$BOD + $\angle$DOF + $\angle$FOA = 180°        (Linear pair)
$\angle$FOA = $\angle$u = $180°-90°-50°=40°$
$\angle \mathrm{FOA}=\angle x=40°$    (Vertically opposite angles)
$\angle \mathrm{BOD}=\angle z=90°$    (Vertically opposite angles)
$\angle \mathrm{EOC}=\angle y=50°$    (Vertically opposite angles)

#### Question 32:

In Fig., find the values of x, y and z.

#### Question 1:

In Fig., line n is a transversal to lines l and m. Identify the following:

(i) Alternate and corresponding angles in Fig. (i).
(ii) Angles alternate to ∠d and ∠g and angles corresponding to angles ∠f and ∠h in Fig. (ii).
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Fig. (iii).
(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. (ii).

(i) Figure (i)
Corresponding angles:
$\angle$EGB and $\angle$GHD
$\angle$HGB and $\angle$FHD
$\angle$EGA and $\angle$GHC
$\angle$AGH and $\angle$CHF
Alternate angles:
$\angle$EGB and $\angle$CHF
$\angle$HGB and $\angle$CHG
$\angle$EGA and $\angle$FHD
$\angle$AGH and $\angle$GHD

(ii) Figure (ii)
Alternate angle to $\angle$d is $\angle$e.
Alternate angle to $\angle$g is $\angle$b.
Also,
Corresponding angle to $\angle$f is $\angle$c.
Corresponding angle to $\angle$h is $\angle$a.

(iii) Figure (iii)
Angle alternate to $\angle$PQR is $\angle$QRA.
Angle corresponding to $\angle$RQF is $\angle$ARB.
Angle alternate to $\angle$POE is $\angle$ARB.

(iv) Figure (ii)
Pair of interior angles are
$\angle$a and $\angle$e
$\angle$d and $\angle$f
Pair of exterior angles are
$\angle$b and $\angle$h
$\angle$c and $\angle$g

#### Question 2:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 60°, find all other angles in the figure.

$\angle$ALM = $\angle$CMQ = $60°$        (Corresponding angles)
$\angle$LMD = $\angle$CMQ = $60°$        (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = $60°$          (Vertically opposite angles)
Since
$\angle$CMQ + $\angle$QMD = $180°$     (Linear pair)
QMD = $180°-60°=120°$
$\angle$QMD = $\angle$MLB = $120°$        (Corresponding angles)
$\angle$QMD = $\angle$CML = $120°$        (Vertically opposite angles)
$\angle$MLB = $\angle$ALP = $120°$          (Vertically opposite angles)

#### Question 3:

In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.

In the given Fig., AB || CD.

#### Question 4:

The line n is transversal to line l and m in Fig. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.

In this given Fig., line l || m.
Here,
Alternate angle to $\angle$13 is $\angle$7.
Corresponding angle to $\angle$15 is $\angle$7.
Alternate angle to $\angle$15 is $\angle$5.

#### Question 5:

In Fig., line l || m and n is a transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.

In the given figure, l || m.
Here,

Also,

Thus,

Hence, alternate angles are equal.

#### Question 6:

In Fig., line l || m and a transversal n cuts them at P and Q respectively. If ∠1 = 75°, find all other angles.

In the given figure, l || m, n is a transversal line and ∠1 = 75°.
Thus, we have:

#### Question 7:

In Fig., AB || CD and a transversal PQ cuts them at L and M respectively. If ∠QMD = 100°, find all other angles.

In the given figure, AB || CD, PQ is a transversal line and $\angle$QMD = 100°.
Thus, we have:
$\angle$DMQ + $\angle$QMC = 180°    (Linear pair)
$\therefore \angle \mathrm{QMC}=180°-\angle \mathrm{DMQ}=180°-100°=80°$
Thus,
$\angle$DMQ = $\angle$BLM = 100°         (Corresponding angles)
$\angle$DMQ = $\angle$CML = 100°         (Vertically opposite angles)
$\angle$BLM = $\angle$PLA = 100°           (Vertically opposite angles)
Also,
$\angle$CMQ = $\angle$ALM = 80°         (Corresponding angles)
$\angle$CMQ = $\angle$DML = 80°         (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = 80°           (Vertically opposite angles)

#### Question 8:

In Fig., l || m and p || q. Find the values of x, y, z, t.

In the given figure, l || m and p || q.
Thus, we have:
$\angle z=80°$                (Vertically opposite angles)
$\angle z=\angle t=80°$       (Corresponding angles)
$\angle z=\angle y=80°$       (Corresponding angles)
$\angle x=\angle y=80°$       (Corresponding angles)

#### Question 9:

In Fig., line l || m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.

In the given figure, ∠1 = 120° and ∠2 =100°.
Since l || m, so

Also,

We know that the sum of all the angles of triangle is 180°.
$\therefore \angle 6+\angle 3+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒60°+80°+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒140°+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒\angle 4=180°-140°=40°$

#### Question 10:

In Fig., line l || m. Find the values of a, b, c, d. Give reasons.

In the given figure, line l || m.
Thus, we have:

#### Question 11:

In Fig., AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

In the given figure, AB || CD and t is a transversal line.
Now, let:
$\angle 1=3x\phantom{\rule{0ex}{0ex}}\angle 2=2x$
Thus, we have:

Now,

#### Question 12:

In Fig., l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.

In the given figure, l || m || n and p is a transversal line.
Thus, we have:

#### Question 13:

In Fig., if l || m || n and ∠1 = 60°, find ∠2.

In the given figure, l || m || n and ∠1 = 60°.
Thus, we have:

#### Question 14:

In Fig., if AB || CD and CD || EF, find ∠ACE.

In the given figure, AB || CD and CD || EF.
Extend line CE to E'.

Thus, we have:

#### Question 15:

In Fig., if l || m, n || p and ∠1 = 85°, find ∠2.

In the given figure, l || m, n || p and ∠1 = 85°.
Now, let ∠4 be the adjacent angle of ∠2.
Thus, we have:

$\angle 3+\angle 2=180°$       (Sum of interior angles on the same side of the transversal)
$\therefore \angle 2=180°-\angle 3=180°-85°=95°$

#### Question 16:

In Fig., a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l || m?

We know that if the alternate exterior angles of two lines are equal, then the lines are parallel.
In the given figure, are alternate exterior angles, but they are not equal.

Therefore, lines l and m are not parallel.

#### Question 17:

In Fig., a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?

$\angle$2 = $\angle$3 = 65°        (Vertically opposite angles)
$\angle$8 = $\angle$6 = 65°         (Vertically opposite angles)
∴ $\angle$3 = $\angle$6
l || m                       (Two lines are parallel if the alternate angles formed with the transversal are equal)

#### Question 18:

In Fig., show that AB || EF.

Extend line CE to E'.

#### Question 19:

In Fig., AB || CD. Find the values of x, y, z.

$\angle x+125°=180°$             (Linear pair)
$\therefore \angle x=180°-125°=55°$

$\angle z=125°$            (Corresponding angles)
$\angle x+\angle z=180°$   (Sum of adjacent interior angles is $180°$)
$\angle x+125°=180°\phantom{\rule{0ex}{0ex}}⇒\angle x=180°-125°=55°$

$\angle x+\angle y=180°$   (Sum of adjacent interior angles is $180°$)
$55°+\angle y=180°\phantom{\rule{0ex}{0ex}}⇒\angle y=180°-55°=125°$

#### Question 20:

In Fig., find out ∠PXR, if PQ || RS.

Draw a line parallel to PQ passing through X.

Here,
(Alternate interior angles)
∵ PQ || RS || XF
∴ $\angle \mathrm{PXR}=\angle \mathrm{PXF}+\angle \mathrm{FXR}=70°+50°=120°$

#### Question 21:

In Fig., we have

(i) ∠MLY = 2 ∠LMQ, find ∠LMQ.
(ii) ∠XLM = (2x − 10)° and ∠LMQ = x + 30°, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x − 15)°, and ∠LMQ = (x + 40)°, find x

(i)

(ii)

(iii)

(iv)

#### Question 22:

In Fig., DE || BC. Find the values of x and y.

$\angle$ABC = $\angle$DAB       (Alternate interior angles)

$\angle$ACB = $\angle$EAC       (Alternate interior angles)

#### Question 23:

In Fig., line AC || line DE and ∠ABD = 32°. Find out the angles x and y if ∠E = 122°.

#### Question 24:

In Fig., side BC of ∆ABC has been produced to D and CE || BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD and ∠ACD.

$\angle$ABC = $\angle$ECD = 55°          (Corresponding angles)
$\angle$BAC = $\angle$ACE = 65°          (Alternate interior angles)
Now, $\angle$ACD = $\angle$ACE + $\angle$ECD
⇒ $\angle$ACD = 55° + 65° = 120°

#### Question 25:

In Fig., line CAAB || line CR and line PR || line BD. Find ∠x, ∠y and ∠z.

Since CA ⊥ AB,
$\therefore \angle x=90°$
We know that the sum of all the angles of triangle is 180°.

$\angle$PBC = $\angle$APQ = $70°$            (Corresponding angles)
Since

#### Question 26:

In Fig., PQ || RS. Find the value of x.

#### Question 27:

In Fig., AB || CD and AE || CF; ∠FCG = 90° and ∠BAC = 120°. Find the values of x, y and z.

$\angle$BAC = $\angle$ACG = 120°          (Alternate interior angle)
∴ $\angle$ACF + $\angle$FCG = 120°
$\angle$ACF = 120° − 90° = 30°

$\angle$DCA + $\angle$ACG = 180°            (Linear pair)
$\angle$x = 180° − 120° = 60°

$\angle$BAC + $\angle$BAE + $\angle$EAC = 360°
$\angle$CAE = 360° − 120° − (60° + 30°) = 150°             ($\angle$BAE =  $\angle$DCF)

#### Question 28:

In Fig., AB || CD and AC || BD. Find the values of x, y, z.

(i) Since AC || BD and CD || AB, ABCD is a parallelogram.
$\angle$CAB + $\angle$ACD = 180°     (Sum of adjacent angles of a parallelogram)
∴ $\angle$ACD = 180° − 65° = 115°
$\angle$CAD = $\angle$CDB = 65°         (Opposite angles of a parallelogram)
$\angle$ACD = $\angle$DBA = 115°       (Opposite angles of a parallelogram)

(ii) Here,
AC || BD and CD || AB
$\angle$DAC = x = 40°            (Alternate interior angle)
$\angle$DAB = y = 35°            (Alternate interior angle)

#### Question 29:

In Fig., state which lines are parallel and why?

Let F be the point of intersection of line CD and the line passing through point E.

Since $\angle$ACD and $\angle$CDE are alternate and equal angles, so
$\angle$ACD = 100° = $\angle$CDE
∴ AC || EF

#### Question 30:

In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.

Construction:
Let G be the point of intersection of lines BC and DE.

∵ AB || DE and BC || EF

$\angle \mathrm{ABC}=\angle \mathrm{DGC}=\angle \mathrm{DEF}=75°$  (Corresponding angles)​

#### Question 1:

The sum of an angle and one third of its supplementary angle is 90°. The measure of the angle is
(a) 135°
(b) 120°
(c) 60°
(d) 45°

Let the required angle be x.
Now, supplementary of the required angle = 180∘ − x
Then,
$x+\frac{1}{3}\left(180°-x\right)=90°\phantom{\rule{0ex}{0ex}}⇒3x+180°-x=270°\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence, the correct answer is option (d).

#### Question 2:

If angles of a linear pair are equal, then the measure of each angle is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Let the required angle be x
Now, Sum of linear pair angles = 180
x + x = 180
2x = 180
x = 90
Hence, the correct answer is option (d).

#### Question 3:

Two complemntary angles are in the ratio 2 : 3. The measure of the larger angle is
(a) 60°
(b) 54°
(c) 66°
(d) 48°

Let the angles be 2x and 3x.
Now, 2x + 3x = 90
⇒ 5x = 90
x = 18
∴ Larger angle = 3x = 3 × 18 = 54
Hence, the correct answer is option (b).

#### Question 4:

An angle is thrice its supplement. The measure of the angle is
(a) 120°
(b) 105°
(c) 135°
(d) 150°

Let the required angle be x
Then,
$x=3\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒x=540°-3x\phantom{\rule{0ex}{0ex}}⇒4x=540°\phantom{\rule{0ex}{0ex}}⇒x=135°$
Hence, the correct answer is option (c).

#### Question 5:

In Fig. 88 PR is a straight line and ∠PQS : ∠SQR = 7 : 5. The measure of ∠SQR is
(a) 60°
(b) ${\left(62\frac{1}{2}\right)}^{°}$
(c) ${\left(67\frac{1}{2}\right)}^{°}$
(d) 75°

Let the measures of the angle ∠PQS and ∠SQR be 7x and 5x.
Now, ∠PQS + ∠SQR = 180             [Linear pair angles]
⇒ 7x + 5x = 180
⇒ 12x = 180
x = 15
∴ ∠SQR = 5x = 5 × 15 = 75
Hence, the correct answer is option (d).

#### Question 6:

The sum of an angle and half of its complementary angle is 75°. The measure of the angle is
(a) 40°
(b) 50°
(c) 60°
(d) 80°

Let the required angle be x.
Now, complementnary of the required angle = 90∘ − x
Then,
$x+\frac{1}{2}\left(90°-x\right)=75°\phantom{\rule{0ex}{0ex}}⇒2x+90°-x=150°\phantom{\rule{0ex}{0ex}}⇒x=150-90°\phantom{\rule{0ex}{0ex}}⇒x=60°$
Hence, the correct answer is option (c).

#### Question 7:

∠A is an obtuse angle. The measure of ∠A and twice its supplementary differ by 30°. Then ∠A can be
(a) 150°
(b) 110°
(c) 140°
(d) 120°

Supplementary of ∠A = 180 − ∠A
Now,
∠A + 30 = 2(180 − ∠A)
⇒ ∠A + 30 = 360 − 2∠A
⇒ 3∠A = 360 − 30
⇒ 3∠A = 330
⇒ ∠A = 110
Hence, the correct answer is option (b).

#### Question 8:

An angle is double of its supplement. The measure of the angle is
(a) 60°
(b) 120°
(c) 40°
(d) 80°

Let the required angle be x.
Now, supplementary of the required angle = 180∘ − x
Then,
$x=2\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒x=360°-2x\phantom{\rule{0ex}{0ex}}⇒3x=360°\phantom{\rule{0ex}{0ex}}⇒x=120°$
Hence, the correct answer is option (b).

#### Question 9:

The measure of an angle which is its own complement is
(a) 30°
(b) 60°
(c) 90°
(d) 45°

Let the required angle be x.
Now, complementary of the required angle = 90∘ − x
Then,
$x=90°-x\phantom{\rule{0ex}{0ex}}⇒x=90°-x\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence, the correct answer is option (d).

#### Question 10:

Two supplementary angles are in the ratio 3 : 2. The smaller angle measures
(a) 108°
(b) 81°
(c) 72°
(d) 68°

Let the angles be 3x and 2x.
Now, 3x + 2x = 180
⇒ 5x = 180
x = 36
∴ Smaller angle = 2x = 2 × 36 = 72
Hence, the correct answer is option (c).

#### Question 11:

In Fig. 89, the value of x is
(a) 75
(b) 65
(c) 45
(d) 55

∠AOC and ∠BOC = 180             [∵ Linear pair angles]
⇒ 44+ (2x + 6) = 180
⇒ (2x + 6) = 136
⇒ 2x + 6 = 136
⇒ 2x = 130
x = 65
Hence, the correct answer is option (b).

#### Question 12:

In Fig. 90, AOB is a straight line and the ray OCstands on it. The value of x is
(a) 16
(b) 26
(c) 36
(d) 46

∠AOC + ∠BOC = 180             [∵ Linear pair angles]
⇒ (2x + 15) + (3x + 35) = 180
⇒ (5x + 50) = 180
⇒ 5x + 50 = 180
⇒ 5x = 130
x = 26
Hence, the correct answer is option (b).

#### Question 13:

In Fig. 91, AOB is a straight line and 4x = 5y. The value of x is
(a) 100
(b) 105
(c) 110
(d) 115

∠AOC + ∠BOC = 180             [∵ Linear pair angles]
y + x = 180
y + x = 180

Hence, the correct answer is option (a).

#### Question 14:

In Fig. 92, AOB is a straight line such that ∠AOC = (3x + 10)°, ∠COD = 50° and ∠BOD = (x − 8)°. The value of x is
(a) 32
(b) 36
(c) 42
(d) 52

∠AOC + ∠COD + ∠BOD = 180             [AOB is a straight line]
⇒ (3x + 10) + 50+ (x − 8) = 180
⇒ 3x + 10 + 50 + x − 8 = 180
⇒ 4x + 52 = 180
⇒ 4x = 128
x = 32
Hence, the correct answer is option (a).

#### Question 15:

In Fig. 93, if AOC is a straight line, then x =
(a) 42°
(b) 52°
(c) 142°
(d) 38°

∠AOD + ∠DOB + ∠BOC = 180             [∵ AOC is a straight line]
⇒ 38 + x + 90 = 180
x + 128 = 180
x = 52
Hence, the correct answer is option (b).

#### Question 16:

In Fig. 94, if  ∠AOC is a straight line, then the value of x is
(a) 15
(b) 18
(c) 20
(d) 16

∠AOD + ∠DOB + ∠BOC = 180             [ AOC is a straight line]
⇒ 2x + 90 + 3x= 180
⇒ 5x + 90 = 180
⇒ 5x = 90
x = 18
Hence, the correct answer is option (b).

#### Question 17:

In Fig. 95, if AB, CD and EF are straight lines, then x =
(a) 5
(b) 10
(c) 20
(d) 30

Let all the lines intersect at O.

∠COF = ∠DOE = 4x                              [Vertically opposite angles]
∠AOC + ∠COF + ∠BOF = 180             [AOB is a straight line]
⇒ 2x + 4x + 3x= 180
⇒ 9x = 180
⇒ 9x = 180
x = 20
Hence, the correct answer is option (c).

#### Question 18:

In Fig. 96, if AB, CD and EF are straight lines, then x + y + z =
(a) 180
(b) 203
(c) 213
(d) 134

∠DAE + ∠BAD + ∠BAF = 180             [EAF is a straight line]
⇒ 3x + 49 + 62= 180
⇒ 3x + 111= 180
⇒ 3x = 69
⇒ 3x = 69
x = 23
Now, ∠CAE + ∠CAF = 180             [∵ EAF is a straight line]
z + y= 180
z + y = 180
Now, x + y + z = 23 + 180 = 203
Hence, the correct answer is option (b).

#### Question 19:

In Fig. 97, if AB is parallel to CD, then the value of ∠BPE is
(a) 106°
(b) 76°
(c) 74°
(d) 84°

Since, AB || CD
∴ ∠BPQ = ∠PQC           [Alternate interior angles]
⇒ (3x + 34) = (5− 14)
⇒ 3x + 34 = 5− 14
⇒ 48 = 2x
x = 24
∴ ∠BPQ = (3 × 24 + 34) = 106
∠BPQ + ∠BPE = 180             [EF is a straight line]
⇒ 106 + ∠BPE = 180
⇒ ∠BPE = 74
Hence, the correct answer is option (c).

#### Question 20:

In Fig. 98, if AB is parallel to CO and EF is a transversal, then x =
(a) 19
(b) 29
(c) 39
(d) 49

Let the line EF intersect AB and CD at P and Q respectively.

Since, AB || CD
∴ ∠BPQ + ∠PQD = 180         (Angles on the same side of a transversal line are supplementary)
⇒ (7x − 12) + (4x + 17) = 180
⇒ 7x − 12 + 4x + 17 = 180
⇒ 11x + 5 = 180
⇒ 11x = 175
x = 15.90

Disclaimer: No option is correct.

#### Question 21:

In Fig. 99, AB || CD and EF is a transversal intersecting ABand CO at Pand Q respectively. The measure of  ∠DPQ is
(a) 100
(b) 80
(c) 110
(d) 70

∠BQF = ∠AQP = (4x)             [Vertically opposite angles]
Since, AB || CD
∴ ∠AQP + ∠CPQ = 180         [Angles on the same side of a transversal line are supplementary]
⇒ (4x) + (5x) = 180
⇒ 9 = 180
x = 20
∴ ∠BQF = (4 × 20) = 80
Now, ∠BQF = ∠DPQ = 80          [Corresponding angles]
Hence, the correct answer is option (b).

#### Question 22:

In Fig. 100, AB || CO and EF is a transversal intersecting AB and CD at P and Q respective. The measure of ∠OOP is
(a) 65
(b) 25
(c) 115
(d) 105

∠BPE = ∠APQ = (5x − 10)        [Vertically opposite angles]
Since, AB || CD
∴ ∠APQ + ∠CQP = 180            [Angles on the same side of a transversal line are supplementary]
⇒ (5x − 10) + (3x − 10) = 180
⇒ 8x − 20  = 180
⇒ 8x = 200
x = 25
∴ ∠BPE = (5 × 25 − 10) = 115
Now, ∠BPE = ∠DQP = 115          [Corresponding angles]
Hence, the correct answer is option (c).

#### Question 23:

In Fig. 101, AB || CD and EF is a transversal. The value of y − x is
(a) 30
(b) 35
(c) 95
(d) 25

Since, AB || CD
∴ ∠BPQ = ∠DQF         [Corresponding angles]
⇒ (5x − 20) = (3x + 40)
⇒ 5x − 20 = 3x + 40
⇒ 2x = 60
x = 30
∴ ∠BPQ = (5 × 30 − 20 ) = 130
Now, ∠APE  = ∠BPQ           [Vertically opposite angles]
⇒ 2y = 130
y = 65
y − x = 65 30 = 35
Hence, the correct answer is option (b).

#### Question 24:

In Fig. 102, AB || CD || EF, ∠ABG = 110°, ∠GCO = 100° and ∠BGC = x°. The value of x is
(a) 35
(b) 50
(c) 30
(d) 40

Since, AB || EG
∴ ∠ABG + ∠EGB = 180         (Angles on the same side of a transversal line are supplementary)
⇒ 110 + ∠EGB = 180
⇒ ∠EGB = 70
Again, CD || GF
∴ ∠DCG + ∠FGC = 180         (Angles on the same side of a transversal line are supplementary)
⇒ 100 + ∠FGC = 180
⇒ ∠FGC = 80
Now, ∠EGB + ∠BGC +∠FGC = 180
⇒ 70 + x + 80= 180
⇒ 150+ x = 180
x = 30
x = 30
Hence, the correct answer is option (c).

#### Question 25:

In Fig. 103, PO || RS and ∠PAB = 60° and ∠ACS = 100°. Then, ∠BAC =
(a) 40°
(b) 60°
(c) 80°
(d) 50°

Since, PQ || RS
∴ ∠PAC = ∠ACS = 100      [Corresponding angles]
Now, ∠PAC = 100
⇒ ∠PAB + ∠BAC = 100
⇒ 60 + ∠BAC = 100
⇒ ∠BAC = 40
Hence, the correct answer is option (a).

#### Question 26:

In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC =
(a) 80°
(b) 90°
(c) 70°
(d) 100°

Construction: Draw a line OE from the point O parallel to AB and CD

Since, AB || OE
∴ ∠BAO + ∠AOE = 180         [Angles on the same side of a transversal line are supplementary]
⇒ 150 + ∠AOE = 180
⇒ ∠AOE = 30
Again, CD || OE
∴ ∠DCO + ∠COE = 180         [Angles on the same side of a transversal line are supplementary]
⇒ 120 + ∠COE = 180
⇒ ∠COE = 60
Now, ∠AOC = ∠AOE + ∠COE
= 30 + 60
= 90
Hence, the correct answer is option (b).

#### Question 27:

In Fig. 105, if AOB and COD are straight lines. Then, x + y =
(a) 120
(b) 140
(c) 100
(d) 160

∠AOD + ∠BOD = 180         [Linear pair angles]
⇒ (7x − 20) + 3x = 180
⇒ 7x − 20 + 3x = 180
⇒ 10x = 200
x = 20
∠AOD = (7 × 20 − 20) = 120
Now∠AOD = ∠BOC = 120                [Vertically opposite angles]
y = 120
Now, x + y = 20 + 120
= 140
Hence, the correct answer is option (b).

#### Question 28:

In Fig. 106, the value of x is
(a) 22
(b) 20
(c) 21
(d) 24

(8x − 41) + (3x) + (3x + 10) + (4x − 5)= 360
⇒ 8x − 41 + 3x + 3x + 10 + 4x − 5 = 360
⇒ 18x − 36 = 360
⇒ 18x = 396
x = 22
Hence, the correct answer is option (a).

#### Question 29:

In Fig. 107, if AOBand COD are straight lines, then
(a) x = 29, y = 100
(b) x = 110, y = 29
(c) x = 29, y = 110
(d) x = 39, y = 110

∠AOD + ∠BOD = 180         [Linear pair angles]
y + 70 = 180
y = 110
y = 110
Now, ∠AOC = ∠BOD = 70                [Vertically opposite angles]
Now, ∠AOC + ∠COE + ∠EOB + ∠BOD + ∠AOD = 360            [Complete angle]
⇒ 70 + 28 + (3x − 5) + 70 + 110 = 360
⇒ (3x) + 273 = 360
⇒ 3x = 87
x = 29
Hence, the correct answer is option (c).

#### Question 30:

In Fig. 108, if AB || CD then the value of x is
(a) 87
(b) 93
(c) 147
(d) 141

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠FCD + Reflex∠FCD = 360         (Complete angle)
⇒ ∠FCD + 273 =  360
⇒ ∠FCD = 87
Since, PQ || CD
∴∠QFC + ∠FCD = 180                 (Angles on the same side of a transversal line are supplementary)
⇒ ∠QFC + 87 = 180
⇒ ∠QFC = 93
Now, ∠ABF = ∠BFQ              (Corresponding angles)
= ∠BFC + ∠QFC
= 54 + 93
= 147
x = 147
x = 147
Hence, the correct answer is option (c).

#### Question 31:

In Fig. 109, if AB || CD then the value of x is
(a) 34
(b) 124
(c) 24
(d) 158

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠QEC + ∠ECD = 180                 [Angles on the same side of a transversal line are supplementary]
⇒ ∠QEC + 56 = 180
⇒ ∠QEC = 124
Now, ∠BEQ + ∠QEC = ∠BEC
⇒ ∠BEQ + 124 = 158
⇒ ∠BEQ = 34
Now, ∠ABE = ∠BEQ = 34              [Corresponding angles]
x = 34
x = 34
Hence, the correct answer is option (a).

#### Question 32:

In Fig. 110, if AB || CD. The value of x is
(a) 122
(b) 238
(c) 58
(d) 119

Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || CD
∴ ∠EFC = ∠FEQ = 37                 [Alternate angles]
Now, ∠AEQ + ∠FEQ = ∠AEF
⇒ ∠AEQ + 37 = 95
⇒ ∠AEQ = 58
Since, PQ || AB
∴∠EAB + ∠AEQ = 180                 [Angles on the same side of a transversal line are supplementary]
⇒ ∠EAB + 58 = 180
⇒ ∠EAB = 122
∠EAB + Reflex∠EAB = 360              [Complete angle]
∴ 122 + (2x) = 360
⇒ 2x = 238
x = 119
Hence, the correct answer is option (d).

#### Question 33:

In Fig. 111, if AB || CO then x =
(a) 154
(b) 139
(c) 144
(d) 164

Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || AB
∴ ∠AME + ∠QEM = 180                [Angles on the same side of a transversal line are supplementary]
⇒ 139 + ∠QEM = 180
⇒ ∠QEM = 41
Now, ∠QEM + ∠DEQ = ∠MED
⇒ 41 + ∠DEQ = 67
⇒ ∠DEQ = 26
Now, ∠PED + ∠DEQ = 180                 [Linear Pair angles]
⇒ ∠PED + 26 = 180
⇒ ∠PED = 154
Since, PQ || AB
x = ∠PED                                         [Corresponding angles]
x = 154
x = 154
Hence, the correct answer is option (a).

#### Question 34:

In Fig. 112, if AB || CD, then x =
(a) 32
(b) 42
(c) 52
(d) 31

Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠CDP + Reflex∠CDP = 360              [Complete angle]
∴∠CDP + 249 = 360
⇒ ∠CDP = 111
Since, PQ || AB
∴ ∠BAP = ∠APQ                                         [Alternate angles]
⇒ ∠BAP = 28
Now, ∠APQ + ∠QPD = ∠APD
⇒ 28 + ∠QPD = (2x + 13)
⇒ ∠QPD = (2x + 13)− 28
Since, PQ || CD
∴ ∠QPD + ∠CDP = 180                [Angles on the same side of a transversal line are supplementary]
⇒ (2x + 13)− 28 + 111 = 180
⇒ 2x + 13 − 28 + 111 = 180
⇒ 2x = 84
x = 42
Hence, the correct answer is option (b).

#### Question 35:

In Fig. 113 if AC || OF and AB || CE, then
(a) x = 145, y = 223
(c) x = 135, y = 233
(b) x = 223, y = 145
(d) x = 233, y = 135

Construction: Produce FD towards D to the point M
∠DCA + Reflex∠DCA = 360              [Complete angle]
∴∠DCA + (y + 15) = 360
⇒ ∠DCA = 345y
Now,
∠MDC = ∠EDF = 58                                    [Vertically Opposite angles]
Since, MF || AC
∴ ∠MDC + ∠QPD = 180                                [Angles on the same side of a transversal line are supplementary]
⇒ 58 + 345y = 180
y = 223
∴ ∠DCA = 345 223 = 122
Again, ∠BAC + Reflex∠BAC = 360              [Complete angle]
∴∠BAC + (2x + 12) = 360
⇒ ∠DCA = 348 − (2x)
Since, AB || CD
∴ ∠DCA + ∠DCA = 180                [Angles on the same side of a transversal line are supplementary]
⇒ 348 − (2x)+ 122 = 180
⇒ (2x)= 290
x = 145
Hence, the correct answer is option (a).

#### Question 1:

Write down each pair of adjacent angles shown in Fig.

Adjacent angles are the angles that have a common vertex and a common arm.
Following are the adjacent angles in the given figure:

#### Question 2:

In Fig., name all the pairs of adjacent angles.

In figure (i), the adjacent angles are:

In figure (ii), the adjacent angles are:

$\angle$BAD and $\angle$DAC
$\angle$BDA and $\angle$CDA

#### Question 3:

In figure, write down: (i) each linear pair (ii) each pair of vertically opposite angles.

(i) Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
$\angle$1 and $\angle$3
$\angle$1 and $\angle$2
$\angle$4 and $\angle$3
$\angle$4 and $\angle$2
$\angle$5 and $\angle$6
$\angle$5 and $\angle$7
$\angle$6 and $\angle$8
$\angle$7 and $\angle$8

(ii) Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
$\angle$1 and $\angle$4
$\angle$2 and $\angle$3
$\angle$5 and $\angle$8
$\angle$6 and $\angle$7

#### Question 4:

Are the angles 1 and 2 given in Fig. adjacent angles?

No, because they have no common vertex.

#### Question 5:

Find the complement of each of the following angles:
(i) 35°
(ii) 72°
(iii) 45°
(iv) 85°

Two angles are called complementary angles if the sum of those angles is 90°.

Complementary angles of the following angles are:

#### Question 6:

Find the supplement of each of the following angles:
(i) 70°
(ii) 120°
(iii) 135°
(iv) 90°

Two angles are called supplementary angles if the sum of those angles is 180°.
Supplementary angles of the following angles are:

(i) 180° − 70° = 110°
(ii) 180° − 120° = 60°
(iii) 180° − 135° = 45°
(iv) 180° − 90° = 90°

#### Question 7:

Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 25°, 65°
(ii) 120°, 60°
(iii) 63°, 27°
(iv) 100°, 80°

Since

Therefore, (i) and (iii) are the pairs of complementary angles and (ii) and (iv) are the pairs of supplementary angles.

#### Question 8:

Can two angles be supplementary, if both of them be
(i) obtuse?
(ii) right?
(iii) acute?

(i) No, two obtuse angles cannot be supplementary.
(ii) Yes, two right angles can be supplementary. ($\because \angle 90°+\angle 90°=\angle 180°$)
(iii) No, two acute angles cannot be supplementary.

#### Question 9:

Name the four pairs of supplementary angles shown in Fig.

Following are the supplementary angles:
$\angle$AOC and $\angle$COB
$\angle$BOC and $\angle$DOB
$\angle$BOD and $\angle$DOA
$\angle$AOC and $\angle$DOA

#### Question 10:

In Fig., A, B, C are collinear points and ∠DBA = ∠EBA.

(i) Name two linear pairs
(ii) Name two pairs of supplementary angles.

(i) Linear pairs:
$\angle$ABD and $\angle$DBC
$\angle$ABE and $\angle$EBC

Because every linear pair forms supplementary angles, these angles are:
$\angle$ABD and $\angle$DBC
$\angle$ABE and $\angle$EBC

#### Question 11:

If two supplementary angles have equal measure, what is the measure of each angle?

Let x and y be two supplementary angles that are equal.
$\angle x=\angle y$
According to the question,
$\angle x+\angle y=180°\phantom{\rule{0ex}{0ex}}⇒\angle x+\angle x=180°\phantom{\rule{0ex}{0ex}}⇒2\angle x=180°\phantom{\rule{0ex}{0ex}}⇒\angle x=\frac{180°}{2}=90°\phantom{\rule{0ex}{0ex}}\therefore \angle x=\angle y=90°$

#### Question 12:

If the complement of an angle is 28°, then find the supplement of the angle.

Let x be the complement of the given angle $28°$.

So, supplement of the angle = $180°-62°=118°$

#### Question 13:

In Fig. 19, name each linear pair and each pair of vertically opposite angles:

Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.

$\angle$1 and $\angle$2
$\angle$2 and $\angle$3
$\angle$3 and $\angle$4
$\angle$1 and $\angle$4
$\angle$5 and $\angle$6
$\angle$6 and $\angle$7
$\angle$7 and $\angle$8
$\angle$8 and $\angle$5
$\angle$9 and $\angle$10
$\angle$10 and $\angle$11
$\angle$11 and $\angle$12
$\angle$12 and $\angle$9

Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
$\angle$1 and $\angle$3
$\angle$4 and $\angle$2
$\angle$5 and $\angle$7
$\angle$6 and $\angle$8
$\angle$9 and $\angle$11
$\angle$10 and $\angle$12

#### Question 14:

In Fig., OE is the bisector of ∠BOD. If ∠1 = 70°, find the magnitudes of ∠2, ∠3 and ∠4.

Since OE is the bisector of $\angle$BOD,

#### Question 15:

One of the angles forming a linear pair is a right angle. What can you say about its other angle?

One angle of a linear pair is the right angle, i.e., 90°.
∴ The other angle = 180°​ 90° = 90​°

#### Question 16:

One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?

If one of the angles of a linear pair is obtuse, then the other angle should be acute; only then can their sum be 180°.

#### Question 17:

One of the angles forming a linear pair is an acute angle. What kind of angle is the other?

In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.

#### Question 18:

Can two acute angles form a linear pair?

No, two acute angles cannot form a linear pair because their sum is always less than 180°.

#### Question 19:

If the supplement of an angle is 65°; then find its complement.

Let be the required angle.
Then, we have:
x + 65° = 180°
$⇒$x = 180° - 65° = 115°

The complement of angle cannot be determined.

#### Question 20:

Find the value of x in each of the following figures.

(i)
Since (Linear pair)

(ii)

(iii)

(iv)

(v)
$2x°+x°+2x°+3x°=180°\phantom{\rule{0ex}{0ex}}⇒8x=180\phantom{\rule{0ex}{0ex}}⇒x=\frac{180}{8}=22.5°$

(vi)
$3x°=105°\phantom{\rule{0ex}{0ex}}⇒x=\frac{105}{3}=35°$

#### Question 21:

In Fig. 22, it being given that ∠1 = 65°, find all other angles.

$\angle 1=\angle 3$          (Vertically opposite angles)
$\therefore \angle 3=65°$
Since $\angle 1+\angle 2=180°$       (Linear pair)
$\therefore \angle 2=180°-65°=115°$
$\angle 2=\angle 4$          (Vertically opposite angles)
$\therefore \angle 4=\angle 2=115°$ and $\angle 3=65°$

#### Question 22:

In Fig., OA and OB are opposite rays:

(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?

$\angle$AOC + $\angle$BOC = 180°                   (Linear pair)
$⇒\left(2y+5\right)+3x=180°\phantom{\rule{0ex}{0ex}}⇒3x+2y=175°$
(i) If x = 25°, then
$3×25°+2y=175°\phantom{\rule{0ex}{0ex}}⇒75°+2y=175°\phantom{\rule{0ex}{0ex}}⇒2y=175°-75°=100°\phantom{\rule{0ex}{0ex}}⇒y=\frac{100°}{2}=50°$
(ii) If y = 35°, then
$3x+2×35°=175°\phantom{\rule{0ex}{0ex}}⇒3x+70°=175°\phantom{\rule{0ex}{0ex}}⇒3x=175°-70°=105°\phantom{\rule{0ex}{0ex}}⇒x=\frac{105°}{3}=35°$

#### Question 23:

In Fig., write all pairs of adjacent angles and all the linear pairs.

Linear pairs of angles:

#### Question 24:

In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.

$\angle BOC=x+20°=50°+20°=70°\phantom{\rule{0ex}{0ex}}\angle COD=x=50°\phantom{\rule{0ex}{0ex}}\angle AOD=x+10°=50°+10°=60°\phantom{\rule{0ex}{0ex}}$

#### Question 25:

How many pairs of adjacent angles are formed when two lines intersect in a point?

If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.

#### Question 26:

How many pairs of adjacent angles, in all, can you name in Fig.?

There are 10 adjacent pairs in the given figure; they are:

#### Question 27:

In Fig., determine the value of x.

#### Question 28:

In Fig., AOC is a line, find x.

#### Question 29:

In Fig., POS is a line, find x.

$\angle \mathrm{QOP}+\angle \mathrm{QOR}+\angle \mathrm{ROS}=180°$       (Angles on a straight line)

$⇒60°+4x+40°=180°\phantom{\rule{0ex}{0ex}}⇒100°+4x=180°\phantom{\rule{0ex}{0ex}}⇒4x=180°-100°=80°\phantom{\rule{0ex}{0ex}}⇒x=\frac{80°}{4}=20°\phantom{\rule{0ex}{0ex}}$

#### Question 30:

In Fig., lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.