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#### Question 1:

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

(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 $180°$.

#### Question 2:

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

(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 $180°$.

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

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

(ii) Complement of $16°=\left(90-16\right)°$$=74°$

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

(iv)

#### Question 3:

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

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

(ii) Complement of $16°=\left(90-16\right)°$$=74°$

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

(iv)

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)
Supplement of

#### 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)
Supplement of

(i) Let the measure of the required angle be $x°$.
Then, in case of complementary angles:
$x+x=90°\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
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:
$x+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, measure of the angle that is equal to its supplement is $90°$.

#### Question 5:

(i) Let the measure of the required angle be $x°$.
Then, in case of complementary angles:
$x+x=90°\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
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:
$x+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, measure of the angle that is equal to its supplement is $90°$.

Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
Therefore,
$x-\left(90°-x\right)=36°\phantom{\rule{0ex}{0ex}}⇒2x=126°\phantom{\rule{0ex}{0ex}}⇒x=63°$
Hence, the measure of the required angle is $63°$.

#### Question 6:

Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
Therefore,
$x-\left(90°-x\right)=36°\phantom{\rule{0ex}{0ex}}⇒2x=126°\phantom{\rule{0ex}{0ex}}⇒x=63°$
Hence, the measure of the required angle is $63°$.

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°.

#### 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°.

Let the measure of the required angle be $x$.
Then, measure of its complement $=\left(90°-x\right)$.
Therefore,
$x=\left(90°-x\right)4\phantom{\rule{0ex}{0ex}}⇒x=360°-4x\phantom{\rule{0ex}{0ex}}⇒5x=360°\phantom{\rule{0ex}{0ex}}⇒x=72°$
Hence, the measure of the required angle is $72°$.

#### Question 8:

Let the measure of the required angle be $x$.
Then, measure of its complement $=\left(90°-x\right)$.
Therefore,
$x=\left(90°-x\right)4\phantom{\rule{0ex}{0ex}}⇒x=360°-4x\phantom{\rule{0ex}{0ex}}⇒5x=360°\phantom{\rule{0ex}{0ex}}⇒x=72°$
Hence, the measure of the required angle is $72°$.

Let the measure of the required angle be $x$.
Then, measure of its supplement $=\left(180°-x\right)$.
Therefore,
$x=\left(180°-x\right)5\phantom{\rule{0ex}{0ex}}⇒x=900°-5x\phantom{\rule{0ex}{0ex}}⇒6x=900°\phantom{\rule{0ex}{0ex}}⇒x=150°$
Hence, the measure of the required angle is $150°$.

#### Question 9:

Let the measure of the required angle be $x$.
Then, measure of its supplement $=\left(180°-x\right)$.
Therefore,
$x=\left(180°-x\right)5\phantom{\rule{0ex}{0ex}}⇒x=900°-5x\phantom{\rule{0ex}{0ex}}⇒6x=900°\phantom{\rule{0ex}{0ex}}⇒x=150°$
Hence, the measure of the required angle is $150°$.

Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
And, measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(180-x\right)=4\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒180-x=360-4x\phantom{\rule{0ex}{0ex}}⇒3x=180\phantom{\rule{0ex}{0ex}}⇒x=60$
Hence, the measure of the required angle is $60°$.

#### Question 10:

Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
And, measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(180-x\right)=4\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒180-x=360-4x\phantom{\rule{0ex}{0ex}}⇒3x=180\phantom{\rule{0ex}{0ex}}⇒x=60$
Hence, the measure of the required angle is $60°$.

Let the measure of the required angle be $x°$.
Then, the measure of its complement $=\left(90-x\right)°$.
And the measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(90-x\right)=\frac{1}{3}\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒3\left(90-x\right)=\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒270-3x=180-x\phantom{\rule{0ex}{0ex}}⇒2x=90\phantom{\rule{0ex}{0ex}}⇒x=45$
Hence, the measure of the required angle is $45°$.

#### Question 11:

Let the measure of the required angle be $x°$.
Then, the measure of its complement $=\left(90-x\right)°$.
And the measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(90-x\right)=\frac{1}{3}\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒3\left(90-x\right)=\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒270-3x=180-x\phantom{\rule{0ex}{0ex}}⇒2x=90\phantom{\rule{0ex}{0ex}}⇒x=45$
Hence, the measure of the required angle is $45°$.

Let the two angles be 4x and 5x, respectively.
Then,
$4x+5x=90\phantom{\rule{0ex}{0ex}}⇒9x=90\phantom{\rule{0ex}{0ex}}⇒x=10°$
Hence, the two angles are .

#### Question 12:

Let the two angles be 4x and 5x, respectively.
Then,
$4x+5x=90\phantom{\rule{0ex}{0ex}}⇒9x=90\phantom{\rule{0ex}{0ex}}⇒x=10°$
Hence, the two angles are .

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° = $\frac{105°}{3}$ = 35°

Thus, the value of x is 35.

#### 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° = $\frac{105°}{3}$ = 35°

Thus, the value of x is 35.

We know that the sum of angles in a linear pair is $180°$.
Therefore,
$\angle AOC+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒62°+x°=180°\phantom{\rule{0ex}{0ex}}⇒x°=\left(180°-62°\right)\phantom{\rule{0ex}{0ex}}⇒x=118°\phantom{\rule{0ex}{0ex}}$
Hence, the value of x is $118°$.

#### Question 2:

We know that the sum of angles in a linear pair is $180°$.
Therefore,
$\angle AOC+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒62°+x°=180°\phantom{\rule{0ex}{0ex}}⇒x°=\left(180°-62°\right)\phantom{\rule{0ex}{0ex}}⇒x=118°\phantom{\rule{0ex}{0ex}}$
Hence, the value of x is $118°$.

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,
$\angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x-7\right)°+55°+\left(x+20\right)°=180\phantom{\rule{0ex}{0ex}}⇒4x=112°\phantom{\rule{0ex}{0ex}}⇒x=28°$
Hence,
$\angle \mathrm{AOC}=3x-7$
$=3×28-7\phantom{\rule{0ex}{0ex}}=77°$
and $\angle \mathrm{BOD}=x+20$
$=28+20\phantom{\rule{0ex}{0ex}}=48°$

#### 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,
$\angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x-7\right)°+55°+\left(x+20\right)°=180\phantom{\rule{0ex}{0ex}}⇒4x=112°\phantom{\rule{0ex}{0ex}}⇒x=28°$
Hence,
$\angle \mathrm{AOC}=3x-7$
$=3×28-7\phantom{\rule{0ex}{0ex}}=77°$
and $\angle \mathrm{BOD}=x+20$
$=28+20\phantom{\rule{0ex}{0ex}}=48°$

AOB is a straight line. Therefore,
$\angle AOC+\angle COD+\angle BOD=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x+7\right)°+\left(2x-19\right)°+x°=180°\phantom{\rule{0ex}{0ex}}⇒6x=192°\phantom{\rule{0ex}{0ex}}⇒x=32°$
Therefore,

#### Question 4:

AOB is a straight line. Therefore,
$\angle AOC+\angle COD+\angle BOD=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x+7\right)°+\left(2x-19\right)°+x°=180°\phantom{\rule{0ex}{0ex}}⇒6x=192°\phantom{\rule{0ex}{0ex}}⇒x=32°$
Therefore,

Let
XOY is a straight line. Therefore,

$\angle XOP+\angle POQ+\angle YOQ=180°\phantom{\rule{0ex}{0ex}}⇒5a+4a+6a=180°\phantom{\rule{0ex}{0ex}}⇒15a=180°\phantom{\rule{0ex}{0ex}}⇒a=12°$
Therefore,

#### Question 5:

Let
XOY is a straight line. Therefore,

$\angle XOP+\angle POQ+\angle YOQ=180°\phantom{\rule{0ex}{0ex}}⇒5a+4a+6a=180°\phantom{\rule{0ex}{0ex}}⇒15a=180°\phantom{\rule{0ex}{0ex}}⇒a=12°$
Therefore,

AOB will be a straight line if
$3x+20+4x-36=180°\phantom{\rule{0ex}{0ex}}⇒7x=196°\phantom{\rule{0ex}{0ex}}⇒x=28°$
Hence, x = 28 will make AOB a straight line.

#### Question 6:

AOB will be a straight line if
$3x+20+4x-36=180°\phantom{\rule{0ex}{0ex}}⇒7x=196°\phantom{\rule{0ex}{0ex}}⇒x=28°$
Hence, x = 28 will make AOB a straight line.

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, $\angle AOC=\angle BOD=50°\phantom{\rule{0ex}{0ex}}$
Let $\angle AOD=\angle BOC=x°$
Also, we know that the sum of all angles around a point is $360°$.
Therefore,
$\angle AOC+\angle AOD+\angle BOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒50+x+50+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=260°\phantom{\rule{0ex}{0ex}}⇒x=130°$
Hence, $\angle AOD=\angle BOC=130°$
Therefore, .

#### Question 7:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, $\angle AOC=\angle BOD=50°\phantom{\rule{0ex}{0ex}}$
Let $\angle AOD=\angle BOC=x°$
Also, we know that the sum of all angles around a point is $360°$.
Therefore,
$\angle AOC+\angle AOD+\angle BOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒50+x+50+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=260°\phantom{\rule{0ex}{0ex}}⇒x=130°$
Hence, $\angle AOD=\angle BOC=130°$
Therefore, .

We know that if two lines intersect, then the vertically opposite angles are equal.
$\therefore \angle BOD=\angle AOC=90°\phantom{\rule{0ex}{0ex}}$
Hence, $t=90°$
Also,
$\angle DOF=\angle COE=50°$
Hence, $z=50°$
Since, AOB is a straight line, we have:
$\angle AOC+\angle COE+\angle BOE=180°\phantom{\rule{0ex}{0ex}}⇒90+50+y=180°\phantom{\rule{0ex}{0ex}}⇒140+y=180°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Also,
$\angle BOE=\angle AOF=40°$
Hence, $x=40°$

#### Question 8:

We know that if two lines intersect, then the vertically opposite angles are equal.
$\therefore \angle BOD=\angle AOC=90°\phantom{\rule{0ex}{0ex}}$
Hence, $t=90°$
Also,
$\angle DOF=\angle COE=50°$
Hence, $z=50°$
Since, AOB is a straight line, we have:
$\angle AOC+\angle COE+\angle BOE=180°\phantom{\rule{0ex}{0ex}}⇒90+50+y=180°\phantom{\rule{0ex}{0ex}}⇒140+y=180°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Also,
$\angle BOE=\angle AOF=40°$
Hence, $x=40°$

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

Since, AOB is a straight line, we have:
$\angle AOE+\angle COE+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒3x+5x+2x=180°\phantom{\rule{0ex}{0ex}}⇒10x=180°\phantom{\rule{0ex}{0ex}}⇒x=18°$

Therefore,
$\angle AOD=2×18°=36°\phantom{\rule{0ex}{0ex}}\angle COE=5×18°=90°\phantom{\rule{0ex}{0ex}}\angle AOE=3×18°=54°$

#### Question 9:

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

Since, AOB is a straight line, we have:
$\angle AOE+\angle COE+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒3x+5x+2x=180°\phantom{\rule{0ex}{0ex}}⇒10x=180°\phantom{\rule{0ex}{0ex}}⇒x=18°$

Therefore,
$\angle AOD=2×18°=36°\phantom{\rule{0ex}{0ex}}\angle COE=5×18°=90°\phantom{\rule{0ex}{0ex}}\angle AOE=3×18°=54°$

Let the two adjacent angles be 5x and 4x, respectively.
Then,
$5x+4x=180°\phantom{\rule{0ex}{0ex}}⇒9x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°$
Hence, the two angles are .

#### Question 10:

Let the two adjacent angles be 5x and 4x, respectively.
Then,
$5x+4x=180°\phantom{\rule{0ex}{0ex}}⇒9x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°$
Hence, the two angles are .

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

And let $\angle BOC=\angle AOD=x$
Also, we know that the sum of all angles around a point is $360°$
$\therefore \angle AOC+\angle BOD+\angle AOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒90°+90°+x+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, $\angle BOC=\angle AOD=90°$
$\therefore \angle AOC=\angle BOD=\angle BOC=\angle AOD=90°$
Hence, the measure of each of the remaining angles is 90o.

#### Question 11:

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

And let $\angle BOC=\angle AOD=x$
Also, we know that the sum of all angles around a point is $360°$
$\therefore \angle AOC+\angle BOD+\angle AOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒90°+90°+x+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, $\angle BOC=\angle AOD=90°$
$\therefore \angle AOC=\angle BOD=\angle BOC=\angle AOD=90°$
Hence, the measure of each of the remaining angles is 90o.

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let $\angle BOC=\angle AOD=x°$
Then,
$x+x=280\phantom{\rule{0ex}{0ex}}⇒2x=280\phantom{\rule{0ex}{0ex}}⇒x=140°\phantom{\rule{0ex}{0ex}}\therefore \angle BOC=\angle AOD=140°$
Also, let $\angle AOC=\angle BOD=y°$
We know that the sum of all angles around a point is $360°$.
$\therefore \angle AOC+\angle BOC+\angle BOD+\angle AOD=360°\phantom{\rule{0ex}{0ex}}⇒y+140+y+140=360°\phantom{\rule{0ex}{0ex}}⇒2y=80°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Hence, $\angle AOC=\angle BOD=40°$

#### Question 12:

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let $\angle BOC=\angle AOD=x°$
Then,
$x+x=280\phantom{\rule{0ex}{0ex}}⇒2x=280\phantom{\rule{0ex}{0ex}}⇒x=140°\phantom{\rule{0ex}{0ex}}\therefore \angle BOC=\angle AOD=140°$
Also, let $\angle AOC=\angle BOD=y°$
We know that the sum of all angles around a point is $360°$.
$\therefore \angle AOC+\angle BOC+\angle BOD+\angle AOD=360°\phantom{\rule{0ex}{0ex}}⇒y+140+y+140=360°\phantom{\rule{0ex}{0ex}}⇒2y=80°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Hence, $\angle AOC=\angle BOD=40°$

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)

#### 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)

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)

#### 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)

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.

#### 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.

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect $\angle AOC$. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let .
We know that vertically-opposite angles are equal.

But, $\angle 1=\angle 2$    [Since OE bisects $\angle AOC$ ]
$\therefore \angle 4=\angle 3$
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

#### Question 16:

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect $\angle AOC$. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let .
We know that vertically-opposite angles are equal.

But, $\angle 1=\angle 2$    [Since OE bisects $\angle AOC$ ]
$\therefore \angle 4=\angle 3$
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Let AOB denote a straight line and let  be the supplementary angles.
Then, we have:

Let .
Then, we have:

Therefore,
$\angle COE+\angle FOC=\frac{1}{2}x+\frac{1}{2}\left(180°-x\right)$
$=\frac{1}{2}\left(x+180°-x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(180°\right)\phantom{\rule{0ex}{0ex}}=90°$

#### Question 1:

Let AOB denote a straight line and let  be the supplementary angles.
Then, we have:

Let .
Then, we have:

Therefore,
$\angle COE+\angle FOC=\frac{1}{2}x+\frac{1}{2}\left(180°-x\right)$
$=\frac{1}{2}\left(x+180°-x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(180°\right)\phantom{\rule{0ex}{0ex}}=90°$

We have, $\angle 1=120°$. Then,

#### Question 2:

We have, $\angle 1=120°$. Then,

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)

#### 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)

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)

#### 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)

For the lines l and m to be parallel

#### Question 5:

For the lines l and m to be parallel

#### Question 6:

$BC\parallel ED$ and CD is the transversal.
Then,

$AB\parallel CD$ and BC is the transversal.

#### Question 7:

$BC\parallel ED$ and CD is the transversal.
Then,

$AB\parallel CD$ and BC is the transversal.

$EF\parallel CD$ and CE is the transversal.
Then,

Again, $AB\parallel CD$ and BC is the transversal.
Then,

#### Question 8:

$EF\parallel CD$ and CE is the transversal.
Then,

Again, $AB\parallel CD$ and BC is the transversal.
Then,

$AB\parallel CD$ and let EF and EG be the transversals.
Now, $AB\parallel CD$  and EF is the transversal.
Then,

Also,

And,

We know that the sum of angles of a triangle is $180°$

#### Question 9:

$AB\parallel CD$ and let EF and EG be the transversals.
Now, $AB\parallel CD$  and EF is the transversal.
Then,

Also,

And,

We know that the sum of angles of a triangle is $180°$

(i)

Draw $EF\parallel AB\parallel CD$.
Now, $AB\parallel EF$ and BE is the transversal.
Then,

Again, $EF\parallel CD$ and DE is the transversal.
Then,

(ii)

Draw $EO\parallel AB\parallel CD$.
Then, $\angle EOB+\angle EOD=x°$
Now, $EO\parallel AB$ and BO is the transversal.

Again, $EO\parallel CD$ and DO is the transversal.

Therefore,

(iii)

Draw $EF\parallel AB\parallel CD$.
Then, $\angle AEF+\angle CEF=x°$
Now, $EF\parallel AB$ and AE is the transversal.

Again, $EF\parallel CD$ and CE is the transversal.

Therefore,

#### Question 10:

(i)

Draw $EF\parallel AB\parallel CD$.
Now, $AB\parallel EF$ and BE is the transversal.
Then,

Again, $EF\parallel CD$ and DE is the transversal.
Then,

(ii)

Draw $EO\parallel AB\parallel CD$.
Then, $\angle EOB+\angle EOD=x°$
Now, $EO\parallel AB$ and BO is the transversal.

Again, $EO\parallel CD$ and DO is the transversal.

Therefore,

(iii)

Draw $EF\parallel AB\parallel CD$.
Then, $\angle AEF+\angle CEF=x°$
Now, $EF\parallel AB$ and AE is the transversal.

Again, $EF\parallel CD$ and CE is the transversal.

Therefore,

Draw $EF\parallel AB\parallel CD$.
$EF\parallel CD$ and CE is the transversal.
Then,

Again, $EF\parallel AB$ and AE is the transversal.
Then,

#### Question 11:

Draw $EF\parallel AB\parallel CD$.
$EF\parallel CD$ and CE is the transversal.
Then,

Again, $EF\parallel AB$ and AE is the transversal.
Then,

Given, $AB\parallel PQ\phantom{\rule{0ex}{0ex}}$.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,

We know that the sum of angles of a triangle is $180°$.
$\therefore \angle GEF+\angle EGF+\angle EFG=180\phantom{\rule{0ex}{0ex}}⇒85°+x+25°=180°\phantom{\rule{0ex}{0ex}}⇒110°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=70°$
And

#### Question 12:

Given, $AB\parallel PQ\phantom{\rule{0ex}{0ex}}$.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,

We know that the sum of angles of a triangle is $180°$.
$\therefore \angle GEF+\angle EGF+\angle EFG=180\phantom{\rule{0ex}{0ex}}⇒85°+x+25°=180°\phantom{\rule{0ex}{0ex}}⇒110°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=70°$
And

$AB\parallel CD$ and AC is the transversal.
Then,

And,

We know that the sum of the angles of a triangle is $180°$.
$\angle ECF+\angle CFE+\angle CEF=180°\phantom{\rule{0ex}{0ex}}⇒105°+30°+x=180°\phantom{\rule{0ex}{0ex}}⇒135°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=45°$

#### Question 13:

$AB\parallel CD$ and AC is the transversal.
Then,

And,

We know that the sum of the angles of a triangle is $180°$.
$\angle ECF+\angle CFE+\angle CEF=180°\phantom{\rule{0ex}{0ex}}⇒105°+30°+x=180°\phantom{\rule{0ex}{0ex}}⇒135°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=45°$

$AB\parallel CD$ and PQ is the transversal.
Then,

And,

Also,

We know that the sum of angles of a triangle is $180°$.
$⇒\angle QGH+\angle GHQ+\angle GQH=180°\phantom{\rule{0ex}{0ex}}⇒95°+65°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°\phantom{\rule{0ex}{0ex}}\therefore x=20°$

#### Question 14:

$AB\parallel CD$ and PQ is the transversal.
Then,

And,

Also,

We know that the sum of angles of a triangle is $180°$.
$⇒\angle QGH+\angle GHQ+\angle GQH=180°\phantom{\rule{0ex}{0ex}}⇒95°+65°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°\phantom{\rule{0ex}{0ex}}\therefore x=20°$

We know that the sum of the angles of a triangle is $180°$.

#### Question 15:

We know that the sum of the angles of a triangle is $180°$.

Draw $EF\parallel AB\parallel CD$ through E.
Now, $EF\parallel AB$ and AE is the transversal.
Then,
Again, $EF\parallel CD$ and CE is the transversal.
Then,

#### Question 16:

Draw $EF\parallel AB\parallel CD$ through E.
Now, $EF\parallel AB$ and AE is the transversal.
Then,
Again, $EF\parallel CD$ and CE is the transversal.
Then,

Draw $PFQ\parallel AB\parallel CD$.
Now, $PFQ\parallel AB$ and EF is the transversal.
Then,

Also, $PFQ\parallel CD$.

#### Question 17:

Draw $PFQ\parallel AB\parallel CD$.
Now, $PFQ\parallel AB$ and EF is the transversal.
Then,

Also, $PFQ\parallel CD$.

In the given figure,

#### Question 18:

In the given figure,

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 = $\frac{1}{2}$∠BEF

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

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

∴ ∠EFG = ∠GFD = $\frac{1}{2}$∠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°

#### 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 = $\frac{1}{2}$∠BEF

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

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

∴ ∠EFG = ∠GFD = $\frac{1}{2}$∠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°

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 = $\frac{1}{2}$∠AEF

⇒ ∠AEF = 2∠FEP          .....(2)

Also, FQ is the bisectors of ∠EFD.

∴ ∠EFQ = ∠QFD = $\frac{1}{2}$∠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)

#### 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 = $\frac{1}{2}$∠AEF

⇒ ∠AEF = 2∠FEP          .....(2)

Also, FQ is the bisectors of ∠EFD.

∴ ∠EFQ = ∠QFD = $\frac{1}{2}$∠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)

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

#### 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

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

â€‹∠ABC + ∠GEF = 180°

Or ∠ABC + ∠DEF = 180°

#### 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

â€‹∠ABC + ∠GEF = 180°

Or ∠ABC + ∠DEF = 180°

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,

â€‹∠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

#### 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,

â€‹∠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

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.

#### 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.

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
$\therefore$
$\angle 1=90°$
Also, â€‹n is perpendicular to
$\therefore$
$\angle 3=90°$
Since p and are parallel and is a transversal line
$\therefore$ $\angle 2=\angle 1=90°$              [Corresponding angles]
Also, $\angle 2=\angle 3=90°$
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.

#### 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
$\therefore$
$\angle 1=90°$
Also, â€‹n is perpendicular to
$\therefore$
$\angle 3=90°$
Since p and are parallel and is a transversal line
$\therefore$ $\angle 2=\angle 1=90°$              [Corresponding angles]
Also, $\angle 2=\angle 3=90°$
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.

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).

#### 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).

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).

#### 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).

(a) acute-angled

Let the angles measure .
Then,
$3x+5x+7x=180°\phantom{\rule{0ex}{0ex}}⇒15x=180°\phantom{\rule{0ex}{0ex}}⇒x=12°$
Therefore, the angles are .
Hence, the triangle is acute-angled.

#### Question 4:

(a) acute-angled

Let the angles measure .
Then,
$3x+5x+7x=180°\phantom{\rule{0ex}{0ex}}⇒15x=180°\phantom{\rule{0ex}{0ex}}⇒x=12°$
Therefore, the angles are .
Hence, the triangle is acute-angled.

Let âˆ†ABC be such that ∠A = 130°.

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

∴ ∠ABP = ∠PBC = $\frac{1}{2}$∠B                .....(1)

Also, ∠ACP = ∠PCB = $\frac{1}{2}$∠C           .....(2)

In âˆ†ABC,

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

⇒ 130° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° − 130° = 50°

$\frac{1}{2}$∠B + $\frac{1}{2}$∠C = $\frac{1}{2}$ × 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).

#### 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 = $\frac{1}{2}$∠B                .....(1)

Also, ∠ACP = ∠PCB = $\frac{1}{2}$∠C           .....(2)

In âˆ†ABC,

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

⇒ 130° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° − 130° = 50°

$\frac{1}{2}$∠B + $\frac{1}{2}$∠C = $\frac{1}{2}$ × 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).

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).

#### 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).

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 âˆ†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).

#### 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 âˆ†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).

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).

#### 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).

(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.

#### 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.

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

Hence, the correct answer is option (d).

#### 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).

(d) 75°

Let the measure of the required angle be $x°$.
Then, the measure of its complement will be $\left(90-x\right)°$.
$\therefore x=5\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒x=450-5x\phantom{\rule{0ex}{0ex}}⇒6x=450\phantom{\rule{0ex}{0ex}}⇒x=75$

#### Question 11:

(d) 75°

Let the measure of the required angle be $x°$.
Then, the measure of its complement will be $\left(90-x\right)°$.
$\therefore x=5\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒x=450-5x\phantom{\rule{0ex}{0ex}}⇒6x=450\phantom{\rule{0ex}{0ex}}⇒x=75$

(b) 54°

Let the measure of the required angle be $x°$.
Then, the measure of its complement will be$\left(90-x\right)°$.
$\mathbf{\therefore }2x=3\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒2x=270-3x\phantom{\rule{0ex}{0ex}}⇒5x=270\phantom{\rule{0ex}{0ex}}⇒x=54$

#### Question 12:

(b) 54°

Let the measure of the required angle be $x°$.
Then, the measure of its complement will be$\left(90-x\right)°$.
$\mathbf{\therefore }2x=3\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒2x=270-3x\phantom{\rule{0ex}{0ex}}⇒5x=270\phantom{\rule{0ex}{0ex}}⇒x=54$

(c) 80°

We have :

(c) 80°

We have :

(b) 86°

We have :

#### Question 14:

(b) 86°

We have :

(c) 80°

We have :

$\therefore \angle AOC={\left[3×30-10\right]}^{°}\phantom{\rule{0ex}{0ex}}⇒\angle AOC=80°$

#### Question 15:

(c) 80°

We have :

$\therefore \angle AOC={\left[3×30-10\right]}^{°}\phantom{\rule{0ex}{0ex}}⇒\angle AOC=80°$

(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.

#### 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.

(b) 30°

Let the measure of the required angle be $x°$
Then, the measure of its supplement will be ${\left(180-x\right)}^{°}$
$\therefore x=\frac{1}{5}\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒5x=180°-x\phantom{\rule{0ex}{0ex}}⇒6x=180°\phantom{\rule{0ex}{0ex}}⇒x=30°$

#### Question 17:

(b) 30°

Let the measure of the required angle be $x°$
Then, the measure of its supplement will be ${\left(180-x\right)}^{°}$
$\therefore x=\frac{1}{5}\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒5x=180°-x\phantom{\rule{0ex}{0ex}}⇒6x=180°\phantom{\rule{0ex}{0ex}}⇒x=30°$

(a) 60°

Let

Then, we have:

(a) 60°

Let

Then, we have:

(c) 45°

We have :

(c) 45°

We have :

(b) 115°

We have :

Now,

#### Question 20:

(b) 115°

We have :

Now,

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let $\angle AQP=\angle BQR=x°$
Now,

#### Question 21:

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let $\angle AQP=\angle BQR=x°$
Now,

(c) 50°

Draw $EOF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

Now,

#### Question 22:

(c) 50°

Draw $EOF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

Now,

(a) 130°

Draw $OE\parallel AB\parallel CD$
Now,  is the transversal.

Also,
is the transversal.

#### Question 23:

(a) 130°

Draw $OE\parallel AB\parallel CD$
Now,  is the transversal.

Also,
is the transversal.

(c) 45°

is the transversal.

Side EC of triangle EFC is produced to D.
$\therefore \angle CEF+\angle EFC=\angle DCF\phantom{\rule{0ex}{0ex}}⇒\angle CEF+25°=80°\phantom{\rule{0ex}{0ex}}⇒\angle CEF=55°$

#### Question 24:

(c) 45°

is the transversal.

Side EC of triangle EFC is produced to D.
$\therefore \angle CEF+\angle EFC=\angle DCF\phantom{\rule{0ex}{0ex}}⇒\angle CEF+25°=80°\phantom{\rule{0ex}{0ex}}⇒\angle CEF=55°$

(b) 126°
Let
Let the transversal intersect AB at P, CD at O and EF at Q.

Then, we have:

Also,
$\angle APO+\angle COP=180°\phantom{\rule{0ex}{0ex}}⇒x+54°=180°\phantom{\rule{0ex}{0ex}}⇒x=126°$

#### Question 25:

(b) 126°
Let
Let the transversal intersect AB at P, CD at O and EF at Q.

Then, we have:

Also,
$\angle APO+\angle COP=180°\phantom{\rule{0ex}{0ex}}⇒x+54°=180°\phantom{\rule{0ex}{0ex}}⇒x=126°$

(a) 50°

is the transversal.

Side QR of traingle PQR is produced to D.
$\therefore \angle PQR+\angle QPR=\angle PRD\phantom{\rule{0ex}{0ex}}⇒70°+\angle QPR=120°\phantom{\rule{0ex}{0ex}}⇒\angle QPR=50°$

#### Question 26:

(a) 50°

is the transversal.

Side QR of traingle PQR is produced to D.
$\therefore \angle PQR+\angle QPR=\angle PRD\phantom{\rule{0ex}{0ex}}⇒70°+\angle QPR=120°\phantom{\rule{0ex}{0ex}}⇒\angle QPR=50°$

(c) 70°

is the transversal.

In $∆ABE$, we have:

#### Question 27:

(c) 70°

is the transversal.

In $∆ABE$, we have:

(c) 30°
In $∆OAB$, we have:

In $∆OCD$, we have:

#### Question 28:

(c) 30°
In $∆OAB$, we have:

In $∆OCD$, we have: