Rs Aggarwal 2021 2022 for Class 9 Maths Chapter 7 - Lines And Angles

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Page No 198:

Answer:

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

(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:

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) $\frac{2}{3} of a right angle \Rightarrow (90 \times \frac{2}{3}) ° = 60 °$
$Complement of \frac{2}{3} of a right angle = (90 - 60) ° = 30 °$

Page No 198:

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) $\frac{3}{5} of a right angle \Rightarrow (\frac{3}{5} \times 90) = 54 °$
Supplement of $\frac{3}{5} of a right angle = (180 - 54) ° = 126 °$

Page No 198:

Answer:

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

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

Page No 198:

Answer:

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

Page No 198:

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:

Answer:

Let the measure of the required angle be $x$ .
Then, measure of its complement $= (90 ° - x)$ .
Therefore,
$x = (90 ° - x) 4 \Rightarrow x = 360 ° - 4 x \Rightarrow 5 x = 360 ° \Rightarrow x = 72 °$
Hence, the measure of the required angle is $72 °$ .

Page No 198:

Answer:

Let the measure of the required angle be $x$ .
Then, measure of its supplement $= (180 ° - x)$ .
Therefore,
$x = (180 ° - x) 5 \Rightarrow x = 900 ° - 5 x \Rightarrow 6 x = 900 ° \Rightarrow x = 150 °$
Hence, the measure of the required angle is $150 °$ .

Page No 198:

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,
$(180 - x) = 4 (90 - x) \Rightarrow 180 - x = 360 - 4 x \Rightarrow 3 x = 180 \Rightarrow x = 60$
Hence, the measure of the required angle is $60 °$ .

Page No 198:

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,
$(90 - x) = \frac{1}{3} (180 - x) \Rightarrow 3 (90 - x) = (180 - x) \Rightarrow 270 - 3 x = 180 - x \Rightarrow 2 x = 90 \Rightarrow x = 45$
Hence, the measure of the required angle is $45 °$ .

Page No 198:

Answer:

Let the two angles be 4x and 5x, respectively.
Then,
$4 x + 5 x = 90 \Rightarrow 9 x = 90 \Rightarrow x = 10 °$
Hence, the two angles are $4 x = 4 \times 10 ° = 40 ° and 5 x = 5 \times 10 ° = 50 °$ .

Page No 198:

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° = $\frac{105 °}{3}$ = 35°

Thus, the value of x is 35.

Page No 206:

Answer:

We know that the sum of angles in a linear pair is $180 °$ .
Therefore,
$∠ A O C + ∠ B O C = 180 ° \Rightarrow 62 ° + x ° = 180 ° \Rightarrow x ° = (180 ° - 62 °) \Rightarrow x = 118 °$
Hence, the value of x is $118 °$ .

Page No 206:

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,
$∠ AOC + ∠ COD + ∠ BOD = 180 ° \Rightarrow (3 x - 7) ° + 55 ° + (x + 20) ° = 180 \Rightarrow 4 x = 112 ° \Rightarrow x = 28 °$
Hence,
$∠ AOC = 3 x - 7$
$= 3 \times 28 - 7 = 77 °$
and $∠ BOD = x + 20$
$= 28 + 20 = 48 °$

Page No 207:

Answer:

AOB is a straight line. Therefore,
$∠ A O C + ∠ C O D + ∠ B O D = 180 ° \Rightarrow (3 x + 7) ° + (2 x - 19) ° + x ° = 180 ° \Rightarrow 6 x = 192 ° \Rightarrow x = 32 °$
Therefore,
$∠ A O C = 3 \times 32 ° + 7 = 103 ° ∠ C O D = 2 \times 32 ° - 19 = 45 ° and ∠ B O D = 32 °$

Page No 207:

Answer:

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

$∠ X O P + ∠ P O Q + ∠ Y O Q = 180 ° \Rightarrow 5 a + 4 a + 6 a = 180 ° \Rightarrow 15 a = 180 ° \Rightarrow a = 12 °$
Therefore,

$x \Rightarrow 5 \times 12 ° = 60 ° y \Rightarrow 4 \times 12 ° = 48 ° and z \Rightarrow 6 \times 12 ° = 72 °$

Page No 207:

Answer:

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

Page No 207:

Answer:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, $∠ A O C = ∠ B O D = 50 °$
Let $∠ A O D = ∠ B O C = x °$
Also, we know that the sum of all angles around a point is $360 °$ .
Therefore,
$∠ A O C + ∠ A O D + ∠ B O D + ∠ B O C = 360 ° \Rightarrow 50 + x + 50 + x = 360 ° \Rightarrow 2 x = 260 ° \Rightarrow x = 130 °$
Hence, $∠ A O D = ∠ B O C = 130 °$
Therefore, $∠ A O D = 130 °, ∠ B O D = 50 ° and ∠ B O C = 130 °$ .

Page No 207:

Answer:

We know that if two lines intersect, then the vertically opposite angles are equal.
$∴ ∠ B O D = ∠ A O C = 90 °$
Hence, $t = 90 °$
Also,
$∠ D O F = ∠ C O E = 50 °$
Hence, $z = 50 °$
Since, AOB is a straight line, we have:
$∠ A O C + ∠ C O E + ∠ B O E = 180 ° \Rightarrow 90 + 50 + y = 180 ° \Rightarrow 140 + y = 180 ° \Rightarrow y = 40 °$
Also,
$∠ B O E = ∠ A O F = 40 °$
Hence, $x = 40 °$
$∴ x = 40 °, y = 40 °, z = 50 ° and t = 90 °$

Page No 207:

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
$∴ ∠ D O F = ∠ C O E = 5 x ° ∠ A O D = ∠ B O C = 2 x ° and ∠ A O E = ∠ B O F = 3 x °$

Since, AOB is a straight line, we have:
$∠ A O E + ∠ C O E + ∠ B O C = 180 ° \Rightarrow 3 x + 5 x + 2 x = 180 ° \Rightarrow 10 x = 180 ° \Rightarrow x = 18 °$

Therefore,
$∠ A O D = 2 \times 18 ° = 36 ° ∠ C O E = 5 \times 18 ° = 90 ° ∠ A O E = 3 \times 18 ° = 54 °$

Page No 207:

Answer:

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

Page No 207:

Answer:

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

$∠ A O C = 90 ° . T hen, ∠ A O C = ∠ B O D = 90 ° .$
And let $∠ B O C = ∠ A O D = x$
Also, we know that the sum of all angles around a point is $360 °$
$∴ ∠ A O C + ∠ B O D + ∠ A O D + ∠ B O C = 360 ° \Rightarrow 90 ° + 90 ° + x + x = 360 ° \Rightarrow 2 x = 180 ° \Rightarrow x = 90 °$
Hence, $∠ B O C = ∠ A O D = 90 °$
$∴ ∠ A O C = ∠ B O D = ∠ B O C = ∠ A O D = 90 °$
Hence, the measure of each of the remaining angles is 90^o.

Page No 208:

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let $∠ B O C = ∠ A O D = x °$
Then,
$x + x = 280 \Rightarrow 2 x = 280 \Rightarrow x = 140 ° ∴ ∠ B O C = ∠ A O D = 140 °$
Also, let $∠ A O C = ∠ B O D = y °$
We know that the sum of all angles around a point is $360 °$ .
$∴ ∠ A O C + ∠ B O C + ∠ B O D + ∠ A O D = 360 ° \Rightarrow y + 140 + y + 140 = 360 ° \Rightarrow 2 y = 80 ° \Rightarrow y = 40 °$
Hence, $∠ A O C = ∠ B O D = 40 °$
$∴ ∠ B O C = ∠ A O D = 140 ° and ∠ A O C = ∠ B O D = 40 °$

Page No 208:

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 = 7k = 7 × 15º = 105º

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

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

Page No 208:

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:

Answer:

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

∴ ∠AOC + ∠BOC = 180º

⇒ xº + 125º = 180º

⇒ xº = 180º − 125º = 55º

Now,

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

∴ yº = 125º

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

∴ zº = 55º

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

Page No 208:

Answer:

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect $∠ A O C$ . Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let $∠ C O E = 1, ∠ A O E = 2, ∠ B O F = 3 and ∠ D O F = 4$ .
We know that vertically-opposite angles are equal.
$∴ ∠ 1 = ∠ 4 and ∠ 2 = ∠ 3$
But, $∠ 1 = ∠ 2$ [Since OE bisects $∠ A O C$ ]
$∴ ∠ 4 = ∠ 3$
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Page No 208:

Answer:

Let AOB denote a straight line and let $∠ A O C and ∠ B O C$ be the supplementary angles.
Then, we have:
$∠ A O C = x ° and ∠ B O C = (180 - x) °$
Let $O E bisect ∠ A O C and O F bisect ∠ B O C$ .
Then, we have:
$∠ A O E = ∠ C O E = \frac{1}{2} x ° and ∠ B O F = ∠ F O C = \frac{1}{2} (180 - x) °$
Therefore,
$∠ C O E + ∠ F O C = \frac{1}{2} x + \frac{1}{2} (180 ° - x)$
$= \frac{1}{2} (x + 180 ° - x) = \frac{1}{2} (180 °) = 90 °$

Page No 223:

Answer:

We have, $∠ 1 = 120 °$ . Then,
$∠ 1 = ∠ 5 [Corresponding angles] \Rightarrow ∠ 5 = 120 ° ∠ 1 = ∠ 3 [Vertically-opposite angles] \Rightarrow ∠ 3 = 120 °$

$∠ 5 = ∠ 7 [Vertically-opposite angles] \Rightarrow ∠ 7 = 120 ° ∠ 1 + ∠ 2 = 180 ° [Since AFB is a straight line] \Rightarrow 120 ° + ∠ 2 = 180 °$

$\Rightarrow ∠ 2 = 60 ° ∠ 2 = ∠ 4 [Vertically-opposite angles] \Rightarrow ∠ 4 = 60 ° ∠ 2 = ∠ 6 [Corresponding angles]$

$\Rightarrow ∠ 6 = 60 ° ∠ 6 = ∠ 8 [Vertically-opposite angles] \Rightarrow ∠ 8 = 60 ° ∴ ∠ 1 = 120 °, ∠ 2 = 60 °, ∠ 3 = 120 °, ∠ 4 = 60 °, ∠ 5 = 120 °, ∠ 6 = 60 °, ∠ 7 = 120 ° and ∠ 8 = 60 °$

Page No 223:

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, l || m and t 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:

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, l || m and t 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:

Answer:

For the lines l and m to be parallel
$\Leftrightarrow 3 x - 20 = 2 x + 10 [Corresponding Angles] \Leftrightarrow x = 30$

Page No 224:

Answer:

$\Leftrightarrow 3 x + 5 + 4 x = 180 [Consecutive Interior Angles] \Leftrightarrow 7 x = 175 \Leftrightarrow x = 25$

Page No 224:

Answer:

$B C ∥ E D$ and CD is the transversal.
Then,

$∠ B C D + ∠ C D E = 180 ° [Angles on the same side of a transversal line are supplementary] \Rightarrow ∠ B C D + 75 = 180 \Rightarrow ∠ B C D = 105 °$

$A B ∥ C D$ and BC is the transversal.

$∠ A B C = ∠ B C D (alternate angles) \Rightarrow x ° = 105 ° \Rightarrow x = 105$

Page No 224:

Answer:

$E F ∥ C D$ and CE is the transversal.
Then,
$∠ E C D + ∠ C E F = 180 ° [Consecutive Interior Angles] \Rightarrow ∠ E C D + 130 ° = 180 ° \Rightarrow ∠ E C D = 50 °$
Again, $A B ∥ C D$ and BC is the transversal.
Then,
$∠ A B C = ∠ B C D [Alternate Interior Angles] \Rightarrow 70 ° = x + 50 ° [∵ ∠ B C D = ∠ B C E + ∠ E C D] \Rightarrow x = 20 °$

Page No 224:

Answer:

$A B ∥ C D$ and let EF and EG be the transversals.
Now, $A B ∥ C D$ and EF is the transversal.
Then,
$∠ A E F = ∠ E F G [Alternate Angles] \Rightarrow y ° = 75 ° \Rightarrow y = 75$
Also,
$∠ E F C + ∠ E F D = 180 ° [Since CFD is a straight line] \Rightarrow x + y = 180 \Rightarrow x + 75 = 180 \Rightarrow x = 105$
And,
$∠ E G F + ∠ E G D = 180 ° [Since CFGD is a straight line] \Rightarrow ∠ E G F + 125 = 180 \Rightarrow ∠ E G F = 55 °$
We know that the sum of angles of a triangle is $180 °$
$∠ E F G + ∠ G E F + ∠ E G F = 180 ° \Rightarrow y + z + 55 = 180 \Rightarrow 75 + z + 55 = 180 \Rightarrow z = 50 ∴ x = 105, y = 75 and z = 50$

Page No 224:

Answer:

(i)

Draw $E F ∥ A B ∥ C D$ .
Now, $A B ∥ E F$ and BE is the transversal.
Then,
$∠ A B E = ∠ B E F [Alternate Interior Angles] \Rightarrow ∠ B E F = 35 °$
Again, $E F ∥ C D$ and DE is the transversal.
Then,
$∠ D E F = ∠ F E D \Rightarrow ∠ F E D = 65 ° ∴ x ° = ∠ B E F + ∠ F E D = (35 + 65) ° = 100 ° or, x = 100$

(ii)

Draw $E O ∥ A B ∥ C D$ .
Then, $∠ E O B + ∠ E O D = x °$
Now, $E O ∥ A B$ and BO is the transversal.
$∴ ∠ E O B + ∠ A B O = 180 ° [Consecutive Interior Angles] \Rightarrow ∠ E O B + 55 ° = 180 ° \Rightarrow ∠ E O B = 125 °$
Again, $E O ∥ C D$ and DO is the transversal.
$∴ ∠ E O D + ∠ C D O = 180 ° [Consecutive Interior Angles] \Rightarrow ∠ E O D + 25 ° = 180 ° \Rightarrow ∠ E O D = 155 °$
Therefore,
$x ° = ∠ E O B + ∠ E O D = (125 + 155) ° = 280 ° or, x = 280$

(iii)

Draw $E F ∥ A B ∥ C D$ .
Then, $∠ A E F + ∠ C E F = x °$
Now, $E F ∥ A B$ and AE is the transversal.
$∴ ∠ A E F + ∠ B A E = 180 ° [Consecutive Interior Angles] \Rightarrow ∠ A E F + 116 = 180 \Rightarrow ∠ A E F = 64 °$
Again, $E F ∥ C D$ and CE is the transversal.
$∠ C E F + ∠ E C D = 180 ° [Consecutive Interior Angles] \Rightarrow ∠ C E F + 124 = 180 \Rightarrow ∠ C E F = 56 °$

Therefore,
$x ° = ∠ A E F + ∠ C E F = (64 + 56) ° = 120 ° or, x = 120$

Page No 225:

Answer:

Draw $E F ∥ A B ∥ C D$ .
$E F ∥ C D$ and CE is the transversal.
Then,
$∠ E C D + ∠ C E F = 180 ° [Angles on the same side of a transversal line are supplementary] \Rightarrow 130 ° + ∠ C E F = 180 ° \Rightarrow ∠ C E F = 50 °$
Again, $E F ∥ A B$ and AE is the transversal.
Then,
$∠ B A E + ∠ A E F = 180 ° [Angles on the same side of a transversal line are supplementary] \Rightarrow x ° + 20 ° + 50 ° = 180 ° [∠ A E F = ∠ A E C + ∠ C E F] \Rightarrow x ° + 70 ° = 180 ° \Rightarrow x ° = 110 ° \Rightarrow x = 110$

Page No 225:

Answer:

Given, $A B ∥ P Q$ .
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
$∠ C E B + ∠ B E G + ∠ G E F = 180 ° [Since CD is a straight line] \Rightarrow 75 ° + 20 ° + ∠ G E F = 180 ° \Rightarrow ∠ G E F = 85 °$
We know that the sum of angles of a triangle is $180 °$ .
$∴ ∠ G E F + ∠ E G F + ∠ E F G = 180 \Rightarrow 85 ° + x + 25 ° = 180 ° \Rightarrow 110 ° + x = 180 ° \Rightarrow x = 70 °$
And
$∠ F E G + ∠ B E G = ∠ D F Q [Corresponding Angles] \Rightarrow 85 ° + 20 ° = ∠ D F Q \Rightarrow ∠ D F Q = 105 ° ∠ E F G + ∠ G F Q + ∠ D F Q = 180 ° [Since CD is a straight line] \Rightarrow 25 ° + y + 105 ° = 180 ° \Rightarrow y = 50 ° ∴ x = 70 ° and y = 50 °$

Page No 225:

Answer:

$A B ∥ C D$ and AC is the transversal.
Then,
$∠ B A C + ∠ A C D = 180 ° [Consecutive Interior Angles] \Rightarrow 75 + ∠ A C D = 180 \Rightarrow ∠ A C D = 105 °$
And,
$∠ A C D = ∠ E C F [Vertically-Opposite Angles] \Rightarrow ∠ E C F = 105 °$
We know that the sum of the angles of a triangle is $180 °$ .
$∠ E C F + ∠ C F E + ∠ C E F = 180 ° \Rightarrow 105 ° + 30 ° + x = 180 ° \Rightarrow 135 ° + x = 180 ° \Rightarrow x = 45 °$

Page No 225:

Answer:

$A B ∥ C D$ and PQ is the transversal.
Then,
$∠ P E F = ∠ E G H [Corresponding Angles] \Rightarrow ∠ E G H = 85 °$
And,
$∠ E G H + ∠ Q G H = 180 ° [Since PQ is a straight line] \Rightarrow 85 ° + ∠ Q G H = 180 ° \Rightarrow ∠ Q G H = 95 °$
Also,
$∠ C H Q + ∠ G H Q = 180 ° [Since CD is a straight line] \Rightarrow 115 ° + ∠ G H Q = 180 ° \Rightarrow ∠ G H Q = 65 °$
We know that the sum of angles of a triangle is $180 °$ .
$\Rightarrow ∠ Q G H + ∠ G H Q + ∠ G Q H = 180 ° \Rightarrow 95 ° + 65 ° + x = 180 ° \Rightarrow x = 20 ° ∴ x = 20 °$

Page No 226:

Answer:

$∠ A D C = ∠ D A B [Alternate Interior Angles] \Rightarrow z = 75 ° ∠ A B C = ∠ B C D [Alternate Interior Angles] \Rightarrow x = 35 °$
We know that the sum of the angles of a triangle is $180 °$ .
$\Rightarrow 35 ° + y + 75 ° = 180 ° \Rightarrow y = 70 ° ∴ x = 35 °, y = 70 ° and z = 75 ° .$

Page No 226:

Answer:

Draw $E F ∥ A B ∥ C D$ through E.
Now, $E F ∥ A B$ and AE is the transversal.
Then, $∠ B A E + ∠ A E F = 180 ° [Angles on the same side of a transversal line are supplementary]$
Again, $E F ∥ C D$ and CE is the transversal.
Then,
$∠ D C E + ∠ C E F = 180 ° [Angles on the same side of a transversal line are supplementary] \Rightarrow ∠ D C E + (∠ A E C + ∠ A E F) = 180 ° \Rightarrow ∠ D C E + ∠ A E C + 180 ° - ∠ B A E = 180 ° \Rightarrow ∠ B A E - ∠ D C E = ∠ A E C$

Page No 226:

Answer:

Draw $P F Q ∥ A B ∥ C D$ .
Now, $P F Q ∥ A B$ and EF is the transversal.
Then,
$∠ A E F + ∠ E F P = 180 ° . . . . . (1) [Angles on the same side of a transversal line are supplementary]$
Also, $P F Q ∥ C D$ .

$∠ P F G = ∠ F G D = r ° [Alternate Angles] and ∠ E F P = ∠ E F G - ∠ P F G = q ° - r ° putting the value of ∠ E F P in eqn . (i) we get, p ° + q ° - r ° = 180 ° \Rightarrow p + q - r = 180$

Page No 226:

Answer:

In the given figure,
$x = 60 ° [Vertically-Opposite Angles] ∠ P R Q = ∠ S Q R [Alternate Angles] y = 60 ° ∠ A P R = ∠ P Q S [Corresponding Angles] \Rightarrow 110 ° = ∠ P Q R + 60 ° [∵ ∠ P Q S = ∠ P Q R + ∠ R Q S] \Rightarrow ∠ P Q R = 50 °$
$∠ P Q R + ∠ R Q S + ∠ B Q S = 180 ° [Since AB is a straight line] \Rightarrow 50 ° + 60 ° + z = 180 ° \Rightarrow 110 ° + z = 180 ° \Rightarrow z = 70 °$
$∠ D S H = z [Corresponding Angles] \Rightarrow ∠ D S H = 70 ° ∴ ∠ D S H = t [Vertically-Opposite Angles] \Rightarrow t = 70 ° ∴ x = 60 °, y = 60 °, z = 70 ° and t = 70 ° .$

Page No 226:

Answer:

It is given that, AB || CD and t 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 ∆EFG,

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

⇒ 90° + ∠EGF = 180° [Using (4)]

⇒ ∠EGF = 180° − 90° = 90°

Page No 227:

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 = $\frac{1}{2}$ ∠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)

Page No 227:

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:

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:

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 ∆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:

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:

Answer:

Let the two lines m and n be respectively perpendicular to the two parallel lines p and q.
To prove: m is parallel to n.
Proof: Since, m is perpendicular to p
$∴$ $∠ 1 = 90 °$
Also, n is perpendicular to q
$∴$ $∠ 3 = 90 °$
Since p and q are parallel and m 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:

Answer:

Let ∆ABC be such that ∠A = ∠B + ∠C.

In ∆ABC,

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

⇒ ∠A + ∠A = 180º (∠A = ∠B + ∠C)

⇒ 2∠A = 180º

⇒ ∠A = 90º

Therefore, ∆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:

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.

∴ x + x = 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:

Answer:

(a) acute-angled

Let the angles measure $(3 x) °, (5 x) ° and (7 x) °$ .
Then,
$3 x + 5 x + 7 x = 180 ° \Rightarrow 15 x = 180 ° \Rightarrow x = 12 °$
Therefore, the angles are $3 (12) ° = 36 °, 5 (12) ° = 60 ° and 7 (12) ° = 84 °$ .
Hence, the triangle is acute-angled.

Page No 232:

Answer:

Let ∆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 ∆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 ∆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:

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:

Answer:

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

Let ∠A = 2k, ∠B = 3k and ∠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:

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 ∆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:

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:

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:

Answer:

(d) 75°

Let the measure of the required angle be $x °$ .
Then, the measure of its complement will be $(90 - x) °$ .
$∴ x = 5 (90 - x) \Rightarrow x = 450 - 5 x \Rightarrow 6 x = 450 \Rightarrow x = 75$

Page No 233:

Answer:

(b) 54°

Let the measure of the required angle be $x °$ .
Then, the measure of its complement will be $(90 - x) °$ .
$∴ 2 x = 3 (90 - x) \Rightarrow 2 x = 270 - 3 x \Rightarrow 5 x = 270 \Rightarrow x = 54$

Page No 233:

Answer:

(c) 80°

We have :
$∠ A O C + ∠ B O C = 180 ° [Since AOB is a straight line] \Rightarrow 4 x + 5 x = 180 ° \Rightarrow 9 x = 180 ° \Rightarrow x = 20 ° ∴ ∠ A O C = 4 \times 20 ° = 80 °$

Page No 233:

Answer:

(b) 86°

We have :
$∠ A O C + ∠ B O C = 180 ° [Since AOB is a straight line] \Rightarrow 3 x + 10 + 4 x - 26 = 180 ° \Rightarrow 7 x = 196 ° \Rightarrow x = 28 °$

$∴ ∠ B O C = {[4 \times 28 - 26]}^{°} Hence, ∠ B O C = 86 ° .$

Page No 233:

Answer:

(c) 80°

We have :
$∠ A O C + ∠ C O D + ∠ B O D = 180 ° [Since AOB is a straight line] \Rightarrow 3 x - 10 + 50 + x + 20 = 180 \Rightarrow 4 x = 120 \Rightarrow x = 30$

$∴ ∠ A O C = {[3 \times 30 - 10]}^{°} \Rightarrow ∠ A O C = 80 °$

Page No 233:

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:

Answer:

(b) 30°

Let the measure of the required angle be $x °$
Then, the measure of its supplement will be ${(180 - x)}^{°}$
$∴ x = \frac{1}{5} (180 ° - x) \Rightarrow 5 x = 180 ° - x \Rightarrow 6 x = 180 ° \Rightarrow x = 30 °$

Page No 233:

Answer:

(a) 60°

Let
$∠ A O B = x ° = (4 a) °, ∠ C O B = y ° = (5 a) ° and ∠ B O D = z ° = (6 a) °$
Then, we have:
$∠ A O B + ∠ C O B + ∠ B O D = 180 ° [Since AOB is a straight line] \Rightarrow 4 a + 5 a + 6 a = 180 ° \Rightarrow 15 a = 180 ° \Rightarrow a = 12 ° ∴ y = 5 \times 12 ° = 60 °$

Page No 233:

Answer:

(c) 45°

We have :
$θ + ϕ = 180 ° [∵ AOD is a straight line] \Rightarrow 3 ϕ + ϕ = 180 ° [∵ θ = 3 ϕ] \Rightarrow 4 ϕ = 180 ° \Rightarrow ϕ = 45 °$

Page No 234:

Answer:

(b) 115°

We have :
$∠ A O C = ∠ B O D [Vertically-Opposite Angles] ∴ ∠ A O C + ∠ B O D = 130 ° \Rightarrow ∠ A O C + ∠ A O C = 130 ° [∵ ∠ A O C = ∠ B O D] \Rightarrow 2 ∠ A O C = 130 ° \Rightarrow ∠ A O C = 65 °$
Now,
$∠ A O C + ∠ A O D = 180 ° [∵ COD is a straight line] \Rightarrow 65 ° + ∠ A O D = 180 ° \Rightarrow ∠ A O C = 115 °$

Page No 234:

Answer:

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let $∠ A Q P = ∠ B Q R = x °$
Now,
$∠ A Q P + ∠ P Q R + ∠ B Q R = 180 ° [∵ AQB is a straight line] \Rightarrow x + 108 + x = 180 ° \Rightarrow 2 x = 72 ° \Rightarrow x = 36 ° ∴ ∠ A Q P = 36 °$

Page No 234:

Answer:

(c) 50°

Draw $E O F ∥ A B ∥ C D$ .
Now, $E O ∥ A B and O A$ is the transversal.
$∴ ∠ E O A = ∠ O A B = 60 ° [Alternate Interior Angles]$
Also,
$O F ∥ C D and O C$ is the transversal.
$∴ ∠ C O F + ∠ O C D = 180 ° [Angles on the same side of a transversal line are supplementary] \Rightarrow ∠ C O F + 110 ° = 180 ° \Rightarrow ∠ C O F = 70 °$
Now,
$∠ E O A + ∠ A O C + ∠ C O F = 180 ° [∵ E O F is a straight line] \Rightarrow 60 ° + ∠ A O C + 70 ° = 180 ° \Rightarrow ∠ A O C = 50 °$

Page No 234:

Answer:

(a) 130°

Draw $O E ∥ A B ∥ C D$
Now, $O E ∥ A B and O A$ is the transversal.
$∴ ∠ O A B + ∠ A O E = 180 ° [Angles on the same side of a transversal line are supplementary] \Rightarrow ∠ O A B + ∠ A O C + ∠ C O E = 180 ° \Rightarrow 100 ° + 30 ° + ∠ C O E = 180 ° \Rightarrow ∠ C O E = 50 °$
Also,
$O E ∥ C D and O C$ is the transversal.
$∴ ∠ O C D + ∠ C O E = 180 ° [Angles on the same side of a transversal line are supplementary] \Rightarrow ∠ O C D + 50 ° = 180 ° \Rightarrow ∠ O C D = 130 °$

Page No 234:

Answer:

(c) 45°

$A B ∥ C D and A F$ is the transversal.
$∴ ∠ D C F = ∠ C A B = 80 ° [Corresponding Angles]$
Side EC of triangle EFC is produced to D.
$∴ ∠ C E F + ∠ E F C = ∠ D C F \Rightarrow ∠ C E F + 25 ° = 80 ° \Rightarrow ∠ C E F = 55 °$

Page No 234:

Answer:

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

Then, we have:
$∠ C O P = ∠ D O F = y [Vertically-Opposite Angles] ∴ ∠ O Q F + ∠ D O Q = 180 ° [Consecutive Interior Angles] \Rightarrow 3 a + 7 a = 180 ° \Rightarrow 10 a = 180 ° \Rightarrow a = 18 ° ∴ y = 3 \times 18 ° = 54 °$
Also,
$∠ A P O + ∠ C O P = 180 ° \Rightarrow x + 54 ° = 180 ° \Rightarrow x = 126 °$

Page No 235:

Answer:

(a) 50°

$A B ∥ C D and P Q$ is the transversal.
$∴ ∠ P Q R = ∠ A P Q = 70 ° [Alternate Interior Angles]$
Side QR of traingle PQR is produced to D.
$∴ ∠ P Q R + ∠ Q P R = ∠ P R D \Rightarrow 70 ° + ∠ Q P R = 120 ° \Rightarrow ∠ Q P R = 50 °$

Page No 235:

Answer:

(c) 70°

$A B ∥ C D and B C$ is the transversal.
$∴ ∠ A B E = ∠ B C D = 60 ° [Alternate Internal Angles]$
In $∆ A B E$ , we have:
$∠ E A B + ∠ A B E + ∠ A E B = 180 ° [Sum of the angles of a triangle] \Rightarrow 50 ° + 60 ° + ∠ A E B = 180 ° \Rightarrow ∠ A E B = 70 °$

Page No 235:

Answer:

(c) 30°
In $∆ O A B$ , we have:
$∠ O A B + ∠ O B A + ∠ A O B = 180 ° [Sum of the angles of a triangle] \Rightarrow 75 ° + 55 ° + ∠ A O B = 180 ° \Rightarrow ∠ A O B = 50 °$
$∴ ∠ C O D = ∠ A O B = 50 ° [Vertically-Opposite Angles]$
In $∆ O C D$ , we have:
$∠ C O D + ∠ O C D + ∠ O D C = 180 ° [Sum of the angles of a triangle] \Rightarrow 50 ° + 100 ° + ∠ O D C = 180 ° \Rightarrow ∠ O D C = 30 °$

Page No 235:

Answer:

(b) 54°

We have:
$3 x + 72 = 180 ° [∵ A O B is a straight line] \Rightarrow 3 x = 108 \Rightarrow x = 36$
Also,
$∠ A O C + ∠ C O D + ∠ B O D = 180 ° [∵ A O B is a straight line] \Rightarrow 36 ° + 90 ° + y = 180 ° \Rightarrow y = 54 °$

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