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Page No 198:

Question 1:

Answer:

(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as AOB.


(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.


(iii) An angle greater than 90° but less than 180° is called an obtuse angle.



(iv) An angle greater than 180° but less than 360° is called a reflex angle.



(v) Two angles are said to be complementary if the sum of their measures is 90°.

(vi) Two angles are said to be supplementary if the sum of their measures is 180°.

Page No 198:

Question 2:

(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as AOB.


(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.


(iii) An angle greater than 90° but less than 180° is called an obtuse angle.



(iv) An angle greater than 180° but less than 360° is called a reflex angle.



(v) Two angles are said to be complementary if the sum of their measures is 90°.

(vi) Two angles are said to be supplementary if the sum of their measures is 180°.

Answer:


Two angles whose sum is 90° are called complementary angles.

(i) Complement of 55° = 90° − 55° = 35°

(ii) Complement of 16°=90-16°=74°

(iii) Complement of 90° = 90° − 90° = 0°

(iv) 23 of a right angle90×23°=60°
Complement of 23 of a right angle=90-60°=30°

Page No 198:

Question 3:


Two angles whose sum is 90° are called complementary angles.

(i) Complement of 55° = 90° − 55° = 35°

(ii) Complement of 16°=90-16°=74°

(iii) Complement of 90° = 90° − 90° = 0°

(iv) 23 of a right angle90×23°=60°
Complement of 23 of a right angle=90-60°=30°

Answer:


Two angles whose sum is 180° are called supplementary angles.

(i) Supplement of 42° = 180° − 42° = 138°

(ii) Supplement of 90° = 180° − 90° = 90°

(iii) Supplement of 124° = 180° − 124° = 56°

(iv) 35 of a right angle35×90=54°
Supplement of 35 of a right angle=180-54°=126°

Page No 198:

Question 4:


Two angles whose sum is 180° are called supplementary angles.

(i) Supplement of 42° = 180° − 42° = 138°

(ii) Supplement of 90° = 180° − 90° = 90°

(iii) Supplement of 124° = 180° − 124° = 56°

(iv) 35 of a right angle35×90=54°
Supplement of 35 of a right angle=180-54°=126°

Answer:

(i) Let the measure of the required angle be x°.
Then, in case of complementary angles:
x+x=90°2x=90°x=45°
Hence, measure of the angle that is equal to its complement is 45°.

(ii) Let the measure of the required angle be x°â€‹.
Then, in case of supplementary angles:
x+x=180°2x=180°x=90°
Hence, measure of the angle that is equal to its supplement is 90°.

Page No 198:

Question 5:

(i) Let the measure of the required angle be x°.
Then, in case of complementary angles:
x+x=90°2x=90°x=45°
Hence, measure of the angle that is equal to its complement is 45°.

(ii) Let the measure of the required angle be x°â€‹.
Then, in case of supplementary angles:
x+x=180°2x=180°x=90°
Hence, measure of the angle that is equal to its supplement is 90°.

Answer:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
Therefore,
x-90°-x=36°2x=126°x=63°
Hence, the measure of the required angle is 63°.

Page No 198:

Question 6:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
Therefore,
x-90°-x=36°2x=126°x=63°
Hence, the measure of the required angle is 63°.

Answer:


Let the measure of the angle be x°.

∴ Supplement of x° = 180° − x°

It is given that,

(180° − x°) − x° = 30°

⇒ 180° − 2x°= 30°

⇒ 2x° = 180° − 30° = 150°

x° = 75°

Thus, the measure of the angle is 75°.

Page No 198:

Question 7:


Let the measure of the angle be x°.

∴ Supplement of x° = 180° − x°

It is given that,

(180° − x°) − x° = 30°

⇒ 180° − 2x°= 30°

⇒ 2x° = 180° − 30° = 150°

x° = 75°

Thus, the measure of the angle is 75°.

Answer:

Let the measure of the required angle be x.
Then, measure of its complement =90°-x.
Therefore,
x=90°-x4x=360°-4x5x=360°x=72°
Hence, the measure of the required angle is 72°.

Page No 198:

Question 8:

Let the measure of the required angle be x.
Then, measure of its complement =90°-x.
Therefore,
x=90°-x4x=360°-4x5x=360°x=72°
Hence, the measure of the required angle is 72°.

Answer:

Let the measure of the required angle be x.
Then, measure of its supplement =180°-x.
Therefore,
x=180°-x5x=900°-5x6x=900°x=150°
Hence, the measure of the required angle is 150°.

Page No 198:

Question 9:

Let the measure of the required angle be x.
Then, measure of its supplement =180°-x.
Therefore,
x=180°-x5x=900°-5x6x=900°x=150°
Hence, the measure of the required angle is 150°.

Answer:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
And, measure of its supplement=180-x°.
Therefore,
180-x=490-x180-x=360-4x3x=180x=60
Hence, the measure of the required angle is 60°.

Page No 198:

Question 10:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
And, measure of its supplement=180-x°.
Therefore,
180-x=490-x180-x=360-4x3x=180x=60
Hence, the measure of the required angle is 60°.

Answer:

Let the measure of the required angle be x°.
Then, the measure of its complement =90-x°.
And the measure of its supplement=180-x°.
Therefore,
90-x=13180-x390-x=180-x270-3x=180-x2x=90x=45
Hence, the measure of the required angle is 45°.

Page No 198:

Question 11:

Let the measure of the required angle be x°.
Then, the measure of its complement =90-x°.
And the measure of its supplement=180-x°.
Therefore,
90-x=13180-x390-x=180-x270-3x=180-x2x=90x=45
Hence, the measure of the required angle is 45°.

Answer:

Let the two angles be 4x and 5x, respectively.
Then,
4x+5x=909x=90x=10°
Hence, the two angles are 4x=4×10°=40° and 5x=5×10°=50°.

Page No 198:

Question 12:

Let the two angles be 4x and 5x, respectively.
Then,
4x+5x=909x=90x=10°
Hence, the two angles are 4x=4×10°=40° and 5x=5×10°=50°.

Answer:


Two angles whose sum is 90° are called complementary angles.

It is given that the angles (2x – 5)° and (x – 10)° are the complementary angles.

∴ (2x – 5)° + (x – 10)° = 90°

⇒ 3x° – 15° = 90°

⇒ 3x° = 90° + 15° = 105°

⇒ x° = 105°3 = 35°

Thus, the value of x is 35.



Page No 206:

Question 1:


Two angles whose sum is 90° are called complementary angles.

It is given that the angles (2x – 5)° and (x – 10)° are the complementary angles.

∴ (2x – 5)° + (x – 10)° = 90°

⇒ 3x° – 15° = 90°

⇒ 3x° = 90° + 15° = 105°

⇒ x° = 105°3 = 35°

Thus, the value of x is 35.

Answer:

We know that the sum of angles in a linear pair is 180°.
Therefore,
AOC+BOC=180°62°+x°=180°x°=180°-62°x=118°
Hence, the value of x is 118°.

Page No 206:

Question 2:

We know that the sum of angles in a linear pair is 180°.
Therefore,
AOC+BOC=180°62°+x°=180°x°=180°-62°x=118°
Hence, the value of x is 118°.

Answer:

As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°.
Therefore,
 AOC+COD+BOD=180°3x-7°+55°+x+20°=1804x=112°x=28°
Hence,
AOC=3x-7
         =3×28-7=77°
and BOD=x+20
                =28+20=48°



Page No 207:

Question 3:

As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°.
Therefore,
 AOC+COD+BOD=180°3x-7°+55°+x+20°=1804x=112°x=28°
Hence,
AOC=3x-7
         =3×28-7=77°
and BOD=x+20
                =28+20=48°

Answer:

AOB is a straight line. Therefore,
AOC+COD+BOD=180°3x+7°+2x-19°+x°=180°6x=192°x=32°
Therefore,
AOC=3×32°+7=103°COD=2×32°-19=45° andBOD=32°

Page No 207:

Question 4:

AOB is a straight line. Therefore,
AOC+COD+BOD=180°3x+7°+2x-19°+x°=180°6x=192°x=32°
Therefore,
AOC=3×32°+7=103°COD=2×32°-19=45° andBOD=32°

Answer:

Let x=5a, y=4a and z=6a
XOY is a straight line. Therefore,

XOP+POQ+YOQ=180°5a+4a+6a=180°15a=180°a=12°
Therefore,

      x5×12°=60°      y4×12°=48°and z6×12°=72°

Page No 207:

Question 5:

Let x=5a, y=4a and z=6a
XOY is a straight line. Therefore,

XOP+POQ+YOQ=180°5a+4a+6a=180°15a=180°a=12°
Therefore,

      x5×12°=60°      y4×12°=48°and z6×12°=72°

Answer:

AOB will be a straight line if
3x+20+4x-36=180°7x=196°x=28°
Hence, x = 28 will make AOB a straight line.

Page No 207:

Question 6:

AOB will be a straight line if
3x+20+4x-36=180°7x=196°x=28°
Hence, x = 28 will make AOB a straight line.

Answer:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, AOC=BOD=50°
Let AOD=BOC=x°
Also, we know that the sum of all angles around a point is 360°.
Therefore, 
AOC+AOD+BOD+BOC=360°50+x+50+x=360°2x=260°x=130°
Hence, AOD=BOC=130°
Therefore, AOD=130°, BOD=50° and BOC=130°.

Page No 207:

Question 7:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, AOC=BOD=50°
Let AOD=BOC=x°
Also, we know that the sum of all angles around a point is 360°.
Therefore, 
AOC+AOD+BOD+BOC=360°50+x+50+x=360°2x=260°x=130°
Hence, AOD=BOC=130°
Therefore, AOD=130°, BOD=50° and BOC=130°.

Answer:

We know that if two lines intersect, then the vertically opposite angles are equal.
BOD=AOC=90°
Hence, t=90°
Also, 
DOF=COE=50°
Hence, z=50°
Since, AOB is a straight line, we have:
AOC+COE+BOE=180°90+50+y=180°140+y=180°y=40°
Also,
BOE=AOF=40°
Hence, x=40°
 x=40°, y=40°, z=50° and t=90°

Page No 207:

Question 8:

We know that if two lines intersect, then the vertically opposite angles are equal.
BOD=AOC=90°
Hence, t=90°
Also, 
DOF=COE=50°
Hence, z=50°
Since, AOB is a straight line, we have:
AOC+COE+BOE=180°90+50+y=180°140+y=180°y=40°
Also,
BOE=AOF=40°
Hence, x=40°
 x=40°, y=40°, z=50° and t=90°

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
DOF=COE=5x°AOD=BOC=2x° andAOE=BOF=3x°

Since, AOB is a straight line, we have:
AOE+COE+BOC=180°3x+5x+2x=180°10x=180°x=18°

Therefore,
AOD=2×18°=36°COE=5×18°=90°AOE=3×18°=54°

Page No 207:

Question 9:

We know that if two lines intersect, then the vertically-opposite angles are equal.
DOF=COE=5x°AOD=BOC=2x° andAOE=BOF=3x°

Since, AOB is a straight line, we have:
AOE+COE+BOC=180°3x+5x+2x=180°10x=180°x=18°

Therefore,
AOD=2×18°=36°COE=5×18°=90°AOE=3×18°=54°

Answer:

Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x+4x=180°9x=180°x=20°
Hence, the two angles are 5×20°=100° and 4×20°=80°.

Page No 207:

Question 10:

Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x+4x=180°9x=180°x=20°
Hence, the two angles are 5×20°=100° and 4×20°=80°.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.

AOC=90°. Then, AOC=BOD=90°.
And let BOC=AOD=x
Also, we know that the sum of all angles around a point is 360°
AOC+BOD+AOD+BOC=360°90°+90°+x+x=360°2x=180°x=90°
Hence, BOC=AOD=90°
AOC=BOD=BOC=AOD=90°
Hence, the measure of each of the remaining angles is 90o.



Page No 208:

Question 11:

We know that if two lines intersect, then the vertically-opposite angles are equal.

AOC=90°. Then, AOC=BOD=90°.
And let BOC=AOD=x
Also, we know that the sum of all angles around a point is 360°
AOC+BOD+AOD+BOC=360°90°+90°+x+x=360°2x=180°x=90°
Hence, BOC=AOD=90°
AOC=BOD=BOC=AOD=90°
Hence, the measure of each of the remaining angles is 90o.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let BOC=AOD=x°
Then,
x+x=2802x=280x=140°BOC=AOD=140°
Also, let AOC=BOD=y°
We know that the sum of all angles around a point is 360°.
AOC+BOC+BOD+AOD=360°y+140+y+140=360°2y=80°y=40°
Hence, AOC=BOD=40°
BOC=AOD=140° and AOC=BOD=40°

Page No 208:

Question 12:

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let BOC=AOD=x°
Then,
x+x=2802x=280x=140°BOC=AOD=140°
Also, let AOC=BOD=y°
We know that the sum of all angles around a point is 360°.
AOC+BOC+BOD+AOD=360°y+140+y+140=360°2y=80°y=40°
Hence, AOC=BOD=40°
BOC=AOD=140° and AOC=BOD=40°

Answer:


Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.

Here, ∠AOC and ∠AOD form a linear pair.

∴ ∠AOC + ∠AOD = 180º

⇒ 5k + 7k = 180º

⇒ 12k = 180º

⇒ k = 15º

∴ ∠AOC = 5k = 5 × 15º = 75º

∠AOD = 7= 7 × 15º = 105º

Now, ∠BOD = ∠AOC = 75º       (Vertically opposite angles)

∠BOC = ∠AOD = 105º       (Vertically opposite angles)

Page No 208:

Question 13:


Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.

Here, ∠AOC and ∠AOD form a linear pair.

∴ ∠AOC + ∠AOD = 180º

⇒ 5k + 7k = 180º

⇒ 12k = 180º

⇒ k = 15º

∴ ∠AOC = 5k = 5 × 15º = 75º

∠AOD = 7= 7 × 15º = 105º

Now, ∠BOD = ∠AOC = 75º       (Vertically opposite angles)

∠BOC = ∠AOD = 105º       (Vertically opposite angles)

Answer:


In the given figure, 

∠AOC = ∠BOD = 40º      (Vertically opposite angles)

∠BOF = ∠AOE = 35º       (Vertically opposite angles)

Now, ∠EOC and ∠COF form a linear pair.

∴ ∠EOC + ∠COF = 180º

⇒ (∠AOE + ∠AOC) + ∠COF = 180º

⇒ 35º + 40º + ∠COF = 180º

⇒ 75º + ∠COF = 180º

⇒ ∠COF = 180º − 75º = 105º

Also,  ∠DOE = ∠COF = 105º       (Vertically opposite angles)

Page No 208:

Question 14:


In the given figure, 

∠AOC = ∠BOD = 40º      (Vertically opposite angles)

∠BOF = ∠AOE = 35º       (Vertically opposite angles)

Now, ∠EOC and ∠COF form a linear pair.

∴ ∠EOC + ∠COF = 180º

⇒ (∠AOE + ∠AOC) + ∠COF = 180º

⇒ 35º + 40º + ∠COF = 180º

⇒ 75º + ∠COF = 180º

⇒ ∠COF = 180º − 75º = 105º

Also,  ∠DOE = ∠COF = 105º       (Vertically opposite angles)

Answer:


Here, ∠AOC and ∠BOC form a linear pair.

∴ ∠AOC + ∠BOC = 180º

⇒ xº + 125º = 180º

⇒ xº = 180º − 125ºâ€‹ = 55ºâ€‹

Now, 

∠AOD = ∠BOC = 125º       (Vertically opposite angles)

∴ yº = 125ºâ€‹

∠BOD = ∠AOC = 55º         (Vertically opposite angles)

∴ zº = 55ºâ€‹

Thus, the respective values of xy and z are 55, 125 and 55.

Page No 208:

Question 15:


Here, ∠AOC and ∠BOC form a linear pair.

∴ ∠AOC + ∠BOC = 180º

⇒ xº + 125º = 180º

⇒ xº = 180º − 125ºâ€‹ = 55ºâ€‹

Now, 

∠AOD = ∠BOC = 125º       (Vertically opposite angles)

∴ yº = 125ºâ€‹

∠BOD = ∠AOC = 55º         (Vertically opposite angles)

∴ zº = 55ºâ€‹

Thus, the respective values of xy and z are 55, 125 and 55.

Answer:

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect AOC. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let COE=1, AOE=2, BOF=3 and DOF=4.
We know that vertically-opposite angles are equal.
1=4 and 2=3
But, 1=2    [Since OE bisects AOC ]
4=3
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Page No 208:

Question 16:

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect AOC. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let COE=1, AOE=2, BOF=3 and DOF=4.
We know that vertically-opposite angles are equal.
1=4 and 2=3
But, 1=2    [Since OE bisects AOC ]
4=3
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Answer:

Let AOB denote a straight line and let AOC and BOC be the supplementary angles.
Then, we have:
AOC=x° and BOC=180-x°
Let OE bisect AOC and OF bisect BOC.
Then, we have:
AOE=COE=12x° andBOF=FOC=12180-x°
Therefore,
COE+FOC=12x+12180°-x
                       =12x+180°-x=12180°=90°



Page No 223:

Question 1:

Let AOB denote a straight line and let AOC and BOC be the supplementary angles.
Then, we have:
AOC=x° and BOC=180-x°
Let OE bisect AOC and OF bisect BOC.
Then, we have:
AOE=COE=12x° andBOF=FOC=12180-x°
Therefore,
COE+FOC=12x+12180°-x
                       =12x+180°-x=12180°=90°

Answer:

We have, 1=120°. Then,
1=5   Corresponding angles5=120°1=3   Vertically-opposite angles3=120°

5=7   Vertically-opposite angles7=120°1+2=180°   Since AFB is a straight line120°+2=180°

2=60°2=4   Vertically-opposite angles4=60°2=6   Corresponding angles

6 =60°6=8   Vertically-opposite angles8=60°1=120°, 2=60°, 3=120°, 4=60°, 5=120°,6 =60°, 7=120° and 8=60°

Page No 223:

Question 2:

We have, 1=120°. Then,
1=5   Corresponding angles5=120°1=3   Vertically-opposite angles3=120°

5=7   Vertically-opposite angles7=120°1+2=180°   Since AFB is a straight line120°+2=180°

2=60°2=4   Vertically-opposite angles4=60°2=6   Corresponding angles

6 =60°6=8   Vertically-opposite angles8=60°1=120°, 2=60°, 3=120°, 4=60°, 5=120°,6 =60°, 7=120° and 8=60°

Answer:


In the given figure, ∠7 and ∠8 form a linear pair.

∴ ∠7 + ∠8 = 180º

⇒ 80º + ∠8 = 180º

⇒ ∠8 = 180º − 80º = 100º

Now, 

∠6 = ∠8 = 100º       (Vertically opposite angles)

∠5 = ∠7 = 80º         (Vertically opposite angles)

It is given that, || m and is a transversal.

∴ ∠1 = ∠5 = 80º      (Pair of corresponding angles)

∠2 = ∠6 = 100º         (Pair of corresponding angles)

∠3 = ∠7 = 80º           (Pair of corresponding angles)

∠4 = ∠8 = 100º         (Pair of corresponding angles)

Page No 223:

Question 3:


In the given figure, ∠7 and ∠8 form a linear pair.

∴ ∠7 + ∠8 = 180º

⇒ 80º + ∠8 = 180º

⇒ ∠8 = 180º − 80º = 100º

Now, 

∠6 = ∠8 = 100º       (Vertically opposite angles)

∠5 = ∠7 = 80º         (Vertically opposite angles)

It is given that, || m and is a transversal.

∴ ∠1 = ∠5 = 80º      (Pair of corresponding angles)

∠2 = ∠6 = 100º         (Pair of corresponding angles)

∠3 = ∠7 = 80º           (Pair of corresponding angles)

∠4 = ∠8 = 100º         (Pair of corresponding angles)

Answer:


Let ∠1 = 2k and ∠2 = 3k, where k is some constant.

Now, ∠1 and ∠2 form a linear pair.

∴ ∠1 + ∠2 = 180º

⇒ 2k + 3k = 180º

⇒ 5k = 180º

k = 36º

∴ ∠1 = 2k = 2 × 36º = 72º

∠2 = 3k = 3 × 36º = 108º

Now, 

∠3 = ∠1 = 72º         (Vertically opposite angles)

∠4 = ∠2 = 108º       (Vertically opposite angles)

It is given that, || m and is a transversal.

∴ ∠5 = ∠1 = 72º       (Pair of corresponding angles)

∠6 = ∠2 = 108º         (Pair of corresponding angles)

∠7 = ∠1 = 72º           (Pair of alternate exterior angles)

∠8 = ∠2 = 108º         (Pair of alternate exterior angles)

Page No 223:

Question 4:


Let ∠1 = 2k and ∠2 = 3k, where k is some constant.

Now, ∠1 and ∠2 form a linear pair.

∴ ∠1 + ∠2 = 180º

⇒ 2k + 3k = 180º

⇒ 5k = 180º

k = 36º

∴ ∠1 = 2k = 2 × 36º = 72º

∠2 = 3k = 3 × 36º = 108º

Now, 

∠3 = ∠1 = 72º         (Vertically opposite angles)

∠4 = ∠2 = 108º       (Vertically opposite angles)

It is given that, || m and is a transversal.

∴ ∠5 = ∠1 = 72º       (Pair of corresponding angles)

∠6 = ∠2 = 108º         (Pair of corresponding angles)

∠7 = ∠1 = 72º           (Pair of alternate exterior angles)

∠8 = ∠2 = 108º         (Pair of alternate exterior angles)

Answer:

For the lines l and m to be parallel
3x-20=2x+10   Corresponding Anglesx=30
 



Page No 224:

Question 5:

For the lines l and m to be parallel
3x-20=2x+10   Corresponding Anglesx=30
 

Answer:

3x+5+4x=180   Consecutive Interior Angles7x=175x=25

Page No 224:

Question 6:

3x+5+4x=180   Consecutive Interior Angles7x=175x=25

Answer:

BCED and CD is the transversal.
Then,

BCD+CDE=180°   Angles on the same side of a transversal line are supplementaryBCD+75=180BCD=105°

ABCD and BC is the transversal.

ABC=BCD  (alternate angles) x°=105°x=105

Page No 224:

Question 7:

BCED and CD is the transversal.
Then,

BCD+CDE=180°   Angles on the same side of a transversal line are supplementaryBCD+75=180BCD=105°

ABCD and BC is the transversal.

ABC=BCD  (alternate angles) x°=105°x=105

Answer:

EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Consecutive Interior AnglesECD+130°=180°ECD=50°
Again, ABCD and BC is the transversal.
Then,
ABC=BCD   Alternate Interior Angles70°=x+50°  BCD=BCE+ECDx=20°

Page No 224:

Question 8:

EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Consecutive Interior AnglesECD+130°=180°ECD=50°
Again, ABCD and BC is the transversal.
Then,
ABC=BCD   Alternate Interior Angles70°=x+50°  BCD=BCE+ECDx=20°

Answer:



ABCD and let EF and EG be the transversals.
Now, ABCD  and EF is the transversal.
Then,
AEF=EFG   Alternate Anglesy°=75°y=75
Also,
EFC+EFD=180°   Since CFD is a straight linex+y=180x+75=180x=105
And,
EGF+EGD=180°   Since CFGD is a straight lineEGF+125=180EGF=55°
We know that the sum of angles of a triangle is 180°
EFG+GEF+EGF=180°y+z+55=18075+z+55=180z=50x=105, y=75 and z=50

Page No 224:

Question 9:



ABCD and let EF and EG be the transversals.
Now, ABCD  and EF is the transversal.
Then,
AEF=EFG   Alternate Anglesy°=75°y=75
Also,
EFC+EFD=180°   Since CFD is a straight linex+y=180x+75=180x=105
And,
EGF+EGD=180°   Since CFGD is a straight lineEGF+125=180EGF=55°
We know that the sum of angles of a triangle is 180°
EFG+GEF+EGF=180°y+z+55=18075+z+55=180z=50x=105, y=75 and z=50

Answer:

(i)

Draw EFABCD.
Now, ABEF and BE is the transversal.
Then,
ABE=BEF   Alternate Interior AnglesBEF=35°
Again, EFCD and DE is the transversal.
Then,
DEF=FEDFED=65°x°=BEF+FED     =35+65°     =100°or, x=100

(ii)

Draw EOABCD.
Then, EOB+EOD=x°
Now, EOAB and BO is the transversal.
EOB+ABO=180°   Consecutive Interior AnglesEOB+55°=180°EOB=125°
Again, EOCD and DO is the transversal.
EOD+CDO=180°   Consecutive Interior AnglesEOD+25°=180°EOD=155°
Therefore,
x°=EOB+EOD =125+155° =280°or, x=280

(iii)

Draw EFABCD.
Then, AEF+CEF=x°
Now, EFAB and AE is the transversal.
 AEF+BAE=180°   Consecutive Interior Angles AEF+116=180AEF=64°
Again, EFCD and CE is the transversal.
CEF+ECD=180°   Consecutive Interior AnglesCEF+124=180CEF=56°

Therefore,
x°=AEF+CEF  =64+56°  =120°or, x=120



Page No 225:

Question 10:

(i)

Draw EFABCD.
Now, ABEF and BE is the transversal.
Then,
ABE=BEF   Alternate Interior AnglesBEF=35°
Again, EFCD and DE is the transversal.
Then,
DEF=FEDFED=65°x°=BEF+FED     =35+65°     =100°or, x=100

(ii)

Draw EOABCD.
Then, EOB+EOD=x°
Now, EOAB and BO is the transversal.
EOB+ABO=180°   Consecutive Interior AnglesEOB+55°=180°EOB=125°
Again, EOCD and DO is the transversal.
EOD+CDO=180°   Consecutive Interior AnglesEOD+25°=180°EOD=155°
Therefore,
x°=EOB+EOD =125+155° =280°or, x=280

(iii)

Draw EFABCD.
Then, AEF+CEF=x°
Now, EFAB and AE is the transversal.
 AEF+BAE=180°   Consecutive Interior Angles AEF+116=180AEF=64°
Again, EFCD and CE is the transversal.
CEF+ECD=180°   Consecutive Interior AnglesCEF+124=180CEF=56°

Therefore,
x°=AEF+CEF  =64+56°  =120°or, x=120

Answer:


Draw EFABCD.
EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Angles on the same side of a transversal line are supplementary130°+CEF=180°CEF=50°
Again, EFAB and AE is the transversal.
Then,
BAE+AEF=180°  Angles on the same side of a transversal line are supplementaryx°+20°+50°=180°   AEF=AEC+CEFx°+70°=180°x°=110°x=110

Page No 225:

Question 11:


Draw EFABCD.
EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Angles on the same side of a transversal line are supplementary130°+CEF=180°CEF=50°
Again, EFAB and AE is the transversal.
Then,
BAE+AEF=180°  Angles on the same side of a transversal line are supplementaryx°+20°+50°=180°   AEF=AEC+CEFx°+70°=180°x°=110°x=110

Answer:



Given, ABPQ.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
CEB+BEG+GEF=180°   Since CD is a straight line75°+20°+GEF=180°GEF=85°
We know that the sum of angles of a triangle is 180°.
GEF+EGF+EFG=18085°+x+25°=180°110°+x=180°x=70°
And
FEG+BEG=DFQ   Corresponding Angles85°+20°=DFQDFQ=105°EFG+GFQ+DFQ=180°   Since CD is a straight line25°+y+105°=180°y=50°x=70° and y=50°

Page No 225:

Question 12:



Given, ABPQ.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
CEB+BEG+GEF=180°   Since CD is a straight line75°+20°+GEF=180°GEF=85°
We know that the sum of angles of a triangle is 180°.
GEF+EGF+EFG=18085°+x+25°=180°110°+x=180°x=70°
And
FEG+BEG=DFQ   Corresponding Angles85°+20°=DFQDFQ=105°EFG+GFQ+DFQ=180°   Since CD is a straight line25°+y+105°=180°y=50°x=70° and y=50°

Answer:

ABCD and AC is the transversal.
Then,
BAC+ACD=180°   Consecutive Interior Angles75+ACD=180ACD=105°
And,
ACD=ECF   Vertically-Opposite AnglesECF=105°
We know that the sum of the angles of a triangle is 180°.
ECF+CFE+CEF=180°105°+30°+x=180°135°+x=180°x=45°

Page No 225:

Question 13:

ABCD and AC is the transversal.
Then,
BAC+ACD=180°   Consecutive Interior Angles75+ACD=180ACD=105°
And,
ACD=ECF   Vertically-Opposite AnglesECF=105°
We know that the sum of the angles of a triangle is 180°.
ECF+CFE+CEF=180°105°+30°+x=180°135°+x=180°x=45°

Answer:

ABCD and PQ is the transversal.
Then,
PEF=EGH   Corresponding AnglesEGH=85°
And,
EGH+QGH=180°   Since PQ is a straight line85°+QGH=180°QGH=95°
Also,
CHQ+GHQ=180°   Since CD is a straight line115°+GHQ=180°GHQ=65°
We know that the sum of angles of a triangle is 180°.
QGH+GHQ+GQH=180°95°+65°+x=180°x=20°x=20°



Page No 226:

Question 14:

ABCD and PQ is the transversal.
Then,
PEF=EGH   Corresponding AnglesEGH=85°
And,
EGH+QGH=180°   Since PQ is a straight line85°+QGH=180°QGH=95°
Also,
CHQ+GHQ=180°   Since CD is a straight line115°+GHQ=180°GHQ=65°
We know that the sum of angles of a triangle is 180°.
QGH+GHQ+GQH=180°95°+65°+x=180°x=20°x=20°

Answer:

ADC=DAB   Alternate Interior Anglesz=75°ABC=BCD   Alternate Interior Anglesx=35°
We know that the sum of the angles of a triangle is 180°.
35°+y+75°=180°y=70°x=35°, y=70° and z=75°.

Page No 226:

Question 15:

ADC=DAB   Alternate Interior Anglesz=75°ABC=BCD   Alternate Interior Anglesx=35°
We know that the sum of the angles of a triangle is 180°.
35°+y+75°=180°y=70°x=35°, y=70° and z=75°.

Answer:


Draw EFABCD through E.
Now, EFAB and AE is the transversal.
Then, BAE+AEF=180°   Angles on the same side of a transversal line are supplementary
Again, EFCD and CE is the transversal.
Then,
DCE+CEF=180°   Angles on the same side of a transversal line are supplementaryDCE+AEC+AEF=180°DCE+AEC+180°-BAE=180°BAE-DCE=AEC

Page No 226:

Question 16:


Draw EFABCD through E.
Now, EFAB and AE is the transversal.
Then, BAE+AEF=180°   Angles on the same side of a transversal line are supplementary
Again, EFCD and CE is the transversal.
Then,
DCE+CEF=180°   Angles on the same side of a transversal line are supplementaryDCE+AEC+AEF=180°DCE+AEC+180°-BAE=180°BAE-DCE=AEC

Answer:


Draw PFQABCD.
Now, PFQAB and EF is the transversal.
Then,
AEF+EFP=180°.....(1)                                                     Angles on the same side of a transversal line are supplementary
Also, PFQCD.

PFG=FGD=r°Alternate  Anglesand EFP=EFG-PFG=q°-r°putting the value of EFP in eqn. (i)we get,p°+q°-r°=180°p+q-r=180

Page No 226:

Question 17:


Draw PFQABCD.
Now, PFQAB and EF is the transversal.
Then,
AEF+EFP=180°.....(1)                                                     Angles on the same side of a transversal line are supplementary
Also, PFQCD.

PFG=FGD=r°Alternate  Anglesand EFP=EFG-PFG=q°-r°putting the value of EFP in eqn. (i)we get,p°+q°-r°=180°p+q-r=180

Answer:

In the given figure,
x=60°   Vertically-Opposite AnglesPRQ=SQR   Alternate Anglesy=60°APR=PQS   Corresponding Angles110°=PQR+60°   PQS=PQR+RQSPQR=50°
PQR+RQS+BQS=180°   Since AB is a straight line50°+60°+z=180°110°+z=180°z=70°
DSH=z   Corresponding AnglesDSH=70°DSH=t   Vertically-Opposite Anglest=70° x=60°, y=60°, z=70° and t=70°.

Page No 226:

Question 18:

In the given figure,
x=60°   Vertically-Opposite AnglesPRQ=SQR   Alternate Anglesy=60°APR=PQS   Corresponding Angles110°=PQR+60°   PQS=PQR+RQSPQR=50°
PQR+RQS+BQS=180°   Since AB is a straight line50°+60°+z=180°110°+z=180°z=70°
DSH=z   Corresponding AnglesDSH=70°DSH=t   Vertically-Opposite Anglest=70° x=60°, y=60°, z=70° and t=70°.

Answer:


It is given that, AB || CD and is a transversal.

 ∠BEF + ∠EFD = 180°     .....(1)     (Sum of the interior angles on the same side of a transversal is supplementary)

EG is the bisector of ∠BEF.    (Given)


∴ ∠BEG = ∠GEF = 12∠BEF

⇒ ∠BEF = 2∠GEF                .....(2)

Also, FG is the bisector of ∠EFD.    (Given)


∴ ∠EFG = ∠GFD = 12∠EFD

⇒ ∠EFD = 2∠EFG                .....(3)

From (1), (2) and (3), we have

2∠GEF + 2∠EFG = 180°

⇒ 2(
∠GEF + ∠EFG) = 180°

⇒ 
∠GEF + ∠EFG = 90°            .....(4)

In âˆ†EFG,

∠GEF + ∠EFG + ∠EGF = 180°          (Angle sum property)

⇒ 90° + ∠EGF = 180°                         [Using (4)]

⇒ ∠EGF = 180° − 90° = 90°



Page No 227:

Question 19:


It is given that, AB || CD and is a transversal.

 ∠BEF + ∠EFD = 180°     .....(1)     (Sum of the interior angles on the same side of a transversal is supplementary)

EG is the bisector of ∠BEF.    (Given)


∴ ∠BEG = ∠GEF = 12∠BEF

⇒ ∠BEF = 2∠GEF                .....(2)

Also, FG is the bisector of ∠EFD.    (Given)


∴ ∠EFG = ∠GFD = 12∠EFD

⇒ ∠EFD = 2∠EFG                .....(3)

From (1), (2) and (3), we have

2∠GEF + 2∠EFG = 180°

⇒ 2(
∠GEF + ∠EFG) = 180°

⇒ 
∠GEF + ∠EFG = 90°            .....(4)

In âˆ†EFG,

∠GEF + ∠EFG + ∠EGF = 180°          (Angle sum property)

⇒ 90° + ∠EGF = 180°                         [Using (4)]

⇒ ∠EGF = 180° − 90° = 90°

Answer:


It is given that, AB || CD and t is a transversal.

∴ ∠AEF = ∠EFD           .....(1)         (Pair of alternate interior angles)

EP is the bisectors of ∠AEF.        (Given)

∴ ∠AEP = ∠FEP = 12∠AEF

⇒ ∠AEF = 2∠FEP          .....(2)

Also, FQ is the bisectors of ∠EFD.

∴ ∠EFQ = ∠QFD = 12∠EFD

⇒ ∠EFD = 2∠EFQ         .....(3)

From (1), (2) and (3), we have

2∠FEP = 2∠EFQ

⇒ ∠FEP = ∠EFQ

Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal. 

∴ EP || FQ        (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Page No 227:

Question 20:


It is given that, AB || CD and t is a transversal.

∴ ∠AEF = ∠EFD           .....(1)         (Pair of alternate interior angles)

EP is the bisectors of ∠AEF.        (Given)

∴ ∠AEP = ∠FEP = 12∠AEF

⇒ ∠AEF = 2∠FEP          .....(2)

Also, FQ is the bisectors of ∠EFD.

∴ ∠EFQ = ∠QFD = 12∠EFD

⇒ ∠EFD = 2∠EFQ         .....(3)

From (1), (2) and (3), we have

2∠FEP = 2∠EFQ

⇒ ∠FEP = ∠EFQ

Thus, the lines EP and FQ are intersected by a transversal EF such that the pair of alternate interior angles formed are equal. 

∴ EP || FQ        (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Answer:


It is given that, BA || ED and BC || EF.

Construction: Extend DE such that it intersects BC at J. Also, extend FE such that it intersects AB at H.



Now, BA || JD and BC is a transversal.

∴ ∠ABC = ∠DJC      .....(1)       (Pair of corresponding angles)

Also, BC || HF and DJ is a transversal.

∴ ∠DJC = ∠DEF      .....(2)       (Pair of corresponding angles)

From (1) and (2), we have

∠ABC = ∠DEF

Page No 227:

Question 21:


It is given that, BA || ED and BC || EF.

Construction: Extend DE such that it intersects BC at J. Also, extend FE such that it intersects AB at H.



Now, BA || JD and BC is a transversal.

∴ ∠ABC = ∠DJC      .....(1)       (Pair of corresponding angles)

Also, BC || HF and DJ is a transversal.

∴ ∠DJC = ∠DEF      .....(2)       (Pair of corresponding angles)

From (1) and (2), we have

∠ABC = ∠DEF

Answer:


It is given that, BA || ED and BC || EF.

Construction: Extend ED such that it intersects BC at G. 



Now, BA || GE and BC is a transversal.

∴ ∠ABC = ∠EGC      .....(1)       (Pair of corresponding angles)

Also, BC || EF and EG is a transversal.

∴ ∠EGC + ∠GEF = 180°      .....(2)       (Interior angles on the same side of the transversal are supplementary)

From (1) and (2), we have

​∠ABC + ∠GEF = 180°        

Or ∠ABC + ∠DEF = 180°          

Page No 227:

Question 22:


It is given that, BA || ED and BC || EF.

Construction: Extend ED such that it intersects BC at G. 



Now, BA || GE and BC is a transversal.

∴ ∠ABC = ∠EGC      .....(1)       (Pair of corresponding angles)

Also, BC || EF and EG is a transversal.

∴ ∠EGC + ∠GEF = 180°      .....(2)       (Interior angles on the same side of the transversal are supplementary)

From (1) and (2), we have

​∠ABC + ∠GEF = 180°        

Or ∠ABC + ∠DEF = 180°          

Answer:


AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.

It is given that the two plane mirrors are perpendicular to each other.

Therefore, BP || OA and AP || OB.

So, BP ⊥ AP       (OA ⊥ OB)

⇒ ∠APB = 90°    .....(1)

In âˆ†APB,

​∠2 + ∠3 + ∠APB = 180°      (Angle sum property)

∴ ∠2 + ∠3 + 90° = 180°       [Using (1)]

⇒ ∠2 + ∠3 = 180° − 90° = 90°

⇒ 2∠2 + 2∠3 = 2 × 90° = 180°        .....(2)

By law of reflection, we have

∠1 = ∠2  and ∠3 = ∠4                     .....(3)       (Angle of incidence = Angle of reflection)   

From (2) and (3), we have

∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ ∠BAC + ∠ABD = 180°            (∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)

Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of the transversal are supplementary.

∴ CA || BD    

Page No 227:

Question 23:


AP is normal to the plane mirror OA and BP is normal to the plane mirror OB.

It is given that the two plane mirrors are perpendicular to each other.

Therefore, BP || OA and AP || OB.

So, BP ⊥ AP       (OA ⊥ OB)

⇒ ∠APB = 90°    .....(1)

In âˆ†APB,

​∠2 + ∠3 + ∠APB = 180°      (Angle sum property)

∴ ∠2 + ∠3 + 90° = 180°       [Using (1)]

⇒ ∠2 + ∠3 = 180° − 90° = 90°

⇒ 2∠2 + 2∠3 = 2 × 90° = 180°        .....(2)

By law of reflection, we have

∠1 = ∠2  and ∠3 = ∠4                     .....(3)       (Angle of incidence = Angle of reflection)   

From (2) and (3), we have

∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ ∠BAC + ∠ABD = 180°            (∠1 + ∠2 = ∠BAC and ∠3 + ∠4 = ∠ABD)

Thus, the lines CA and BD are intersected by a transversal AB such that the interior angles on the same side of the transversal are supplementary.

∴ CA || BD    

Answer:


Here, ∠BAC = ∠ACD = 110°

Thus, lines AB aand CD are intersected by a transversal AC such that the pair of alternate angles are equal.

∴ AB || CD     (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Thus, line AB is parallel to line CD.

Also, ∠ACD + ∠CDE = 110° + 80° = 190° ≠ 180°

If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.

Therefore, line AC is not parallel to line DE.      



Page No 228:

Question 24:


Here, ∠BAC = ∠ACD = 110°

Thus, lines AB aand CD are intersected by a transversal AC such that the pair of alternate angles are equal.

∴ AB || CD     (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel)

Thus, line AB is parallel to line CD.

Also, ∠ACD + ∠CDE = 110° + 80° = 190° ≠ 180°

If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.

Therefore, line AC is not parallel to line DE.      

Answer:



Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: is parallel to n.
Proof: Since, m is perpendicular to p
1=90°
Also, ​n is perpendicular to 
3=90°
Since p and are parallel and is a transversal line 
 2=1=90°              [Corresponding angles]
Also, 2=3=90°
We know that if two corresponding angles are equal then the two lines containing them must be parallel.
Therefore, the lines m and n are parallel to each other.



Page No 231:

Question 1:



Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: is parallel to n.
Proof: Since, m is perpendicular to p
1=90°
Also, ​n is perpendicular to 
3=90°
Since p and are parallel and is a transversal line 
 2=1=90°              [Corresponding angles]
Also, 2=3=90°
We know that if two corresponding angles are equal then the two lines containing them must be parallel.
Therefore, the lines m and n are parallel to each other.

Answer:


Let âˆ†ABC be such that ∠A = ∠B + ∠C.

In âˆ†ABC,

∠A + ∠B + ∠C = 180º    (Angle sum property)

⇒ ∠A + ∠A = 180º         (∠A = ∠B + ∠C)

⇒ 2∠A = 180º

⇒ ∠A = 90º

Therefore, âˆ†ABC is a right triangle.

Thus, if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.

Hence, the correct answer is option (d).



Page No 232:

Question 2:


Let âˆ†ABC be such that ∠A = ∠B + ∠C.

In âˆ†ABC,

∠A + ∠B + ∠C = 180º    (Angle sum property)

⇒ ∠A + ∠A = 180º         (∠A = ∠B + ∠C)

⇒ 2∠A = 180º

⇒ ∠A = 90º

Therefore, âˆ†ABC is a right triangle.

Thus, if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.

Hence, the correct answer is option (d).

Answer:


Let the measure of each of the two equal interior opposite angles of the triangle be x.

In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.

∴ xx = 110°

⇒ 2x = 110°

⇒ x = 55°

Thus, the measure of each of these equal angles is 55°.

Hence, the correct answer is option (b).

Page No 232:

Question 3:


Let the measure of each of the two equal interior opposite angles of the triangle be x.

In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.

∴ xx = 110°

⇒ 2x = 110°

⇒ x = 55°

Thus, the measure of each of these equal angles is 55°.

Hence, the correct answer is option (b).

Answer:

(a) acute-angled

Let the angles measure 3x°, 5x° and 7x°.
Then,
3x+5x+7x=180°15x=180°x=12°
Therefore, the angles are 312°=36°, 512°=60° and 712°=84°.
Hence, the triangle is acute-angled.

Page No 232:

Question 4:

(a) acute-angled

Let the angles measure 3x°, 5x° and 7x°.
Then,
3x+5x+7x=180°15x=180°x=12°
Therefore, the angles are 312°=36°, 512°=60° and 712°=84°.
Hence, the triangle is acute-angled.

Answer:


Let âˆ†ABC be such that ∠A = 130°.



Here, BP is the bisector of ∠B and CP is the bisector of ∠C.

∴ ∠ABP = ∠PBC = 12∠B                .....(1)

Also, ∠ACP = ∠PCB = 12∠C           .....(2)

In âˆ†ABC,

∠A + ∠B + ∠C = 180°         (Angle sum property)

⇒ 130° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° − 130° = 50°

12∠B + 12∠C = 12 × 50° = 25°

⇒ ∠PBC + ∠PCB = 25°     .....(3)     [Using (1) and (2)]

In âˆ†PBC,

∠PBC + ∠PCB + ∠BPC = 180°         (Angle sum property)

⇒ 25° + ∠BPC = 180°                       [Using (3)]

⇒ ∠BPC = 180° − 25° = 155°

Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.

Hence, the correct answer is option (d).

Page No 232:

Question 5:


Let âˆ†ABC be such that ∠A = 130°.



Here, BP is the bisector of ∠B and CP is the bisector of ∠C.

∴ ∠ABP = ∠PBC = 12∠B                .....(1)

Also, ∠ACP = ∠PCB = 12∠C           .....(2)

In âˆ†ABC,

∠A + ∠B + ∠C = 180°         (Angle sum property)

⇒ 130° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° − 130° = 50°

12∠B + 12∠C = 12 × 50° = 25°

⇒ ∠PBC + ∠PCB = 25°     .....(3)     [Using (1) and (2)]

In âˆ†PBC,

∠PBC + ∠PCB + ∠BPC = 180°         (Angle sum property)

⇒ 25° + ∠BPC = 180°                       [Using (3)]

⇒ ∠BPC = 180° − 25° = 155°

Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.

Hence, the correct answer is option (d).

Answer:


It is given that, AOB is a straight line.

∴ 60º + (5xº + 3xº) = 180º           (Linear pair)

⇒ 8xº = 180º − 60º = 120º

⇒ xº = 15º

Thus, the value of x is 15.

Hence, the correct answer is option (b).

Page No 232:

Question 6:


It is given that, AOB is a straight line.

∴ 60º + (5xº + 3xº) = 180º           (Linear pair)

⇒ 8xº = 180º − 60º = 120º

⇒ xº = 15º

Thus, the value of x is 15.

Hence, the correct answer is option (b).

Answer:


Suppose âˆ†ABC be such that ∠A : ∠B : ∠C = 2 : 3 : 4.

Let ∠A = 2k, ∠B = 3and ∠C = 4k, where k is some constant.

In ∆ABC,

∠A + ∠B + ∠C = 180º      (Angle sum property)

⇒ 2k + 3k + 4k = 180º

⇒ 9k = 180º

⇒ k = 20º

∴ Measure of the largest angle = 4k = 4 × 20º = 80º

Hence, the correct answer is option (c).

Page No 232:

Question 7:


Suppose âˆ†ABC be such that ∠A : ∠B : ∠C = 2 : 3 : 4.

Let ∠A = 2k, ∠B = 3and ∠C = 4k, where k is some constant.

In ∆ABC,

∠A + ∠B + ∠C = 180º      (Angle sum property)

⇒ 2k + 3k + 4k = 180º

⇒ 9k = 180º

⇒ k = 20º

∴ Measure of the largest angle = 4k = 4 × 20º = 80º

Hence, the correct answer is option (c).

Answer:


In the given figure, OA || CD.

Construction: Extend OA such that it intersects BC at E.



Now, OE || CD and BC is a transversal.

∴ ∠AEC = ∠BCD = 130°       (Pair of corresponding angles)

Also, ∠OAB + ∠BAE = 180°         (Linear pair)

∴ 110° + ∠BAE = 180°

⇒ ∠BAE = 180° − 110° = 70°

In âˆ†ABE,

∠AEC = ∠BAE + ∠ABE                (In a triangle, exterior angle is equal to the sum of two opposite interior angles)

∴ 130° = 70° + x°

⇒ x° = 130° − 70° = 60°

Thus, the measure of angle ∠ABC is 60°.

Hence, the correct answer is option (c).

Page No 232:

Question 8:


In the given figure, OA || CD.

Construction: Extend OA such that it intersects BC at E.



Now, OE || CD and BC is a transversal.

∴ ∠AEC = ∠BCD = 130°       (Pair of corresponding angles)

Also, ∠OAB + ∠BAE = 180°         (Linear pair)

∴ 110° + ∠BAE = 180°

⇒ ∠BAE = 180° − 110° = 70°

In âˆ†ABE,

∠AEC = ∠BAE + ∠ABE                (In a triangle, exterior angle is equal to the sum of two opposite interior angles)

∴ 130° = 70° + x°

⇒ x° = 130° − 70° = 60°

Thus, the measure of angle ∠ABC is 60°.

Hence, the correct answer is option (c).

Answer:

(a) an acute angle

If two angles are complements of each other, that is, the sum of their measures is 90°, then each angle is an acute angle.

Page No 232:

Question 9:

(a) an acute angle

If two angles are complements of each other, that is, the sum of their measures is 90°, then each angle is an acute angle.

Answer:


An angle which measures more than 180° but less than 360° is called a reflex angle.

Hence, the correct answer is option (d).

Page No 232:

Question 10:


An angle which measures more than 180° but less than 360° is called a reflex angle.

Hence, the correct answer is option (d).

Answer:

(d) 75°

Let the measure of the required angle be x°.
Then, the measure of its complement will be 90-x°.
x=590-xx=450-5x6x=450x=75



Page No 233:

Question 11:

(d) 75°

Let the measure of the required angle be x°.
Then, the measure of its complement will be 90-x°.
x=590-xx=450-5x6x=450x=75

Answer:

(b) 54°

Let the measure of the required angle be x°.
Then, the measure of its complement will be90-x°.
2x=390-x2x=270-3x5x=270x=54

Page No 233:

Question 12:

(b) 54°

Let the measure of the required angle be x°.
Then, the measure of its complement will be90-x°.
2x=390-x2x=270-3x5x=270x=54

Answer:

(c) 80°

We have :
AOC+BOC=180°   Since AOB is a straight line4x+5x=180°9x=180°x=20°AOC=4×20°=80°

Page No 233:

Question 13:

(c) 80°

We have :
AOC+BOC=180°   Since AOB is a straight line4x+5x=180°9x=180°x=20°AOC=4×20°=80°

Answer:

(b) 86°

We have :
AOC+BOC=180°   Since AOB is a straight line 3x+10+4x-26=180°7x=196°x=28°

BOC=4×28-26°Hence, BOC=86°.

Page No 233:

Question 14:

(b) 86°

We have :
AOC+BOC=180°   Since AOB is a straight line 3x+10+4x-26=180°7x=196°x=28°

BOC=4×28-26°Hence, BOC=86°.

Answer:

(c) 80°

We have :
AOC+COD+BOD=180°   Since AOB is a straight line 3x-10+50+x+20=1804x=120x=30

AOC=3×30-10°AOC=80°

Page No 233:

Question 15:

(c) 80°

We have :
AOC+COD+BOD=180°   Since AOB is a straight line 3x-10+50+x+20=1804x=120x=30

AOC=3×30-10°AOC=80°

Answer:

(a) Through a given point, only one straight line can be drawn.

Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.

Page No 233:

Question 16:

(a) Through a given point, only one straight line can be drawn.

Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.

Answer:

(b) 30°

Let the measure of the required angle be x°
Then, the measure of its supplement will be 180-x°
x=15180°-x5x=180°-x6x=180°x=30°

Page No 233:

Question 17:

(b) 30°

Let the measure of the required angle be x°
Then, the measure of its supplement will be 180-x°
x=15180°-x5x=180°-x6x=180°x=30°

Answer:

(a) 60°

Let  
AOB=x°=4a°, COB=y°=5a° and BOD=z°=6a°
Then, we have:
AOB+COB+BOD=180°   Since AOB is a straight line4a+5a+6a=180° 15a=180°a=12°y=5×12°=60°

Page No 233:

Question 18:

(a) 60°

Let  
AOB=x°=4a°, COB=y°=5a° and BOD=z°=6a°
Then, we have:
AOB+COB+BOD=180°   Since AOB is a straight line4a+5a+6a=180° 15a=180°a=12°y=5×12°=60°

Answer:

(c) 45°

We have :
θ+ϕ=180°   AOD is a straight line3ϕ+ϕ=180°   θ=3ϕ4ϕ=180°ϕ=45°



Page No 234:

Question 19:

(c) 45°

We have :
θ+ϕ=180°   AOD is a straight line3ϕ+ϕ=180°   θ=3ϕ4ϕ=180°ϕ=45°

Answer:

(b) 115°

We have :
AOC=BOD   Vertically-Opposite AnglesAOC+BOD=130°AOC+AOC=130°   AOC=BOD2AOC=130°AOC=65°
Now,
AOC+AOD=180°  COD is a straight line 65°+AOD=180°AOC=115°

Page No 234:

Question 20:

(b) 115°

We have :
AOC=BOD   Vertically-Opposite AnglesAOC+BOD=130°AOC+AOC=130°   AOC=BOD2AOC=130°AOC=65°
Now,
AOC+AOD=180°  COD is a straight line 65°+AOD=180°AOC=115°

Answer:

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let AQP=BQR=x°
Now,
AQP+PQR+BQR=180°   AQB is a straight line x+108+x=180°2x=72°x=36°AQP=36°

Page No 234:

Question 21:

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let AQP=BQR=x°
Now,
AQP+PQR+BQR=180°   AQB is a straight line x+108+x=180°2x=72°x=36°AQP=36°

Answer:

(c) 50°

Draw EOFABCD.
Now, EOAB and OA is the transversal.
EOA=OAB=60°  Alternate Interior Angles 
Also,
OFCD and OC is the transversal.
COF+OCD=180°   Angles on the same side of a transversal line are supplementaryCOF+110°=180°COF=70°
Now,
EOA+AOC+COF=180°   EOF is a straight line60°+AOC+70°=180°AOC=50°

Page No 234:

Question 22:

(c) 50°

Draw EOFABCD.
Now, EOAB and OA is the transversal.
EOA=OAB=60°  Alternate Interior Angles 
Also,
OFCD and OC is the transversal.
COF+OCD=180°   Angles on the same side of a transversal line are supplementaryCOF+110°=180°COF=70°
Now,
EOA+AOC+COF=180°   EOF is a straight line60°+AOC+70°=180°AOC=50°

Answer:

(a) 130°

Draw OEABCD
Now, OEAB and OA is the transversal.
OAB+AOE=180°   Angles on the same side of a transversal line are supplementaryOAB+AOC+COE=180°100°+30°+COE=180°COE=50°
Also, 
OECD and OC is the transversal.
OCD+COE=180°   Angles on the same side of a transversal line are supplementaryOCD+50°=180°OCD=130°

Page No 234:

Question 23:

(a) 130°

Draw OEABCD
Now, OEAB and OA is the transversal.
OAB+AOE=180°   Angles on the same side of a transversal line are supplementaryOAB+AOC+COE=180°100°+30°+COE=180°COE=50°
Also, 
OECD and OC is the transversal.
OCD+COE=180°   Angles on the same side of a transversal line are supplementaryOCD+50°=180°OCD=130°

Answer:

(c) 45°

ABCD and AF is the transversal.
DCF=CAB=80°   Corresponding Angles
Side EC of triangle EFC is produced to D.
CEF+EFC=DCFCEF+25°=80°CEF=55°

Page No 234:

Question 24:

(c) 45°

ABCD and AF is the transversal.
DCF=CAB=80°   Corresponding Angles
Side EC of triangle EFC is produced to D.
CEF+EFC=DCFCEF+25°=80°CEF=55°

Answer:

(b) 126°
Let y=3a° and z=7a°
Let the transversal intersect AB at P, CD at O and EF at Q.

Then, we have:
COP=DOF=y   Vertically-Opposite AnglesOQF+DOQ=180°   Consecutive Interior Angles3a+7a=180°10a=180°a=18°y=3×18°=54°
Also,
APO+COP=180°x+54°=180°x=126°



Page No 235:

Question 25:

(b) 126°
Let y=3a° and z=7a°
Let the transversal intersect AB at P, CD at O and EF at Q.

Then, we have:
COP=DOF=y   Vertically-Opposite AnglesOQF+DOQ=180°   Consecutive Interior Angles3a+7a=180°10a=180°a=18°y=3×18°=54°
Also,
APO+COP=180°x+54°=180°x=126°

Answer:

(a) 50°

ABCD and PQ is the transversal.
PQR=APQ=70°   Alternate Interior Angles
Side QR of traingle PQR is produced to D.
PQR+QPR=PRD70°+QPR=120°QPR=50°

Page No 235:

Question 26:

(a) 50°

ABCD and PQ is the transversal.
PQR=APQ=70°   Alternate Interior Angles
Side QR of traingle PQR is produced to D.
PQR+QPR=PRD70°+QPR=120°QPR=50°

Answer:

(c) 70°

ABCD and BC is the transversal.
ABE=BCD=60°   Alternate Internal Angles
In ABE, we have:
EAB+ABE+AEB=180°   Sum of the angles of a triangle50°+60°+AEB=180°AEB=70°

Page No 235:

Question 27:

(c) 70°

ABCD and BC is the transversal.
ABE=BCD=60°   Alternate Internal Angles
In ABE, we have:
EAB+ABE+AEB=180°   Sum of the angles of a triangle50°+60°+AEB=180°AEB=70°

Answer:

(c) 30°
In OAB, we have:
OAB+OBA+AOB=180°   Sum of the angles of a triangle75°+55°+AOB=180°AOB=50°
COD=AOB=50°   Vertically-Opposite Angles
In OCD, we have:
COD+OCD+ODC=180°   Sum of the angles of a triangle50°+100°+ODC=180°ODC=30°

Page No 235:

Question 28:

(c) 30°
In OAB, we have:
OAB+OBA+AOB=180°   Sum of the angles of a triangle75°+55°+AOB=180°AOB=50°
COD=AOB=50°   Vertically-Opposite Angles
In OCD, we have:
COD+OCD+ODC=180°   Sum of the angles of a triangle50°+100°+ODC=180°ODC=30°

Answer:

(b) 54°

We have:
3x+72=180°   AOB is a straight line3x=108x=36
Also,
AOC+COD+BOD=180°   AOB is a straight line36°+90°+y=180°y=54°



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