Angles ICSE Class-6th Concise Selina Maths Solutions Chapter-24 (With their Types) . We provide step by step Solutions of Exercise / lesson-24 Angles (With their Types) for ICSE Class-6 Concise Selina Mathematics.

Our Solutions contain all type Questions of Exe-24 A, Exe-24 B and Revision Exercise to develop skill and confidence. Visit official Website  for detail information about ICSE Board Class-6

## Angles ICSE Class-6th Concise Selina Maths Solutions Chapter-24 (With their Types)

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Exe-24 A,

Exe-24 B,

Revision Exercise

### Exercise – 24 AAngles (With their Types) for ICSE Class-6th Concise Selina Maths Solutions

#### Question -1.

For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.

(i) In figure (i) O is the vertex, OA, OB are its arms and name of the angle is ∠AOB or∠BOA or simply ∠O.
(ii) In figure (ii) Q is the vertex, QP and QR its arms and the name of the angle is ∠PQR or ∠RQP or simply ∠Q.
(iii) In figure (iii), M is the vertex, MN and ML and its anus, and name of the angle is ∠LMN or ∠NML or simply ∠M.

#### Question -2.

Name the points :
(i) in the interior of the angle PQR,
(ii) in the exterior of the angle PQR.

(i) a, b and x
(ii) d, m, n, s, and t.

#### Question- 3.

In the given figure, figure out the number of angles formed within the arms OA and OE.

∠AOE, ∠AOD, ∠AOC, ∠AOB, ∠BOC ∠BOD, ∠BOE, ∠COD, ∠COE and ∠DOE

#### Question- 4.

(i) 29° 16′ 23″ and 8° 27′ 12″
(ii) 9° 45’56” and 73° 8′ 15″
(m) 56° 38′ and 27° 42’30”
(iv) 47° and 61° 17’4″

(i) 29° 16′ 23″ and 8° 27′ 12″

29° 16′ 23″

+ 8° 27′ 12″

37° 43′ 35”

(ii) 9° 45’56” and 73° 8′ 15″

9° 45’56”

+ 73° 8′ 15″
82° 53′ 71″

Since, 60” = 1′ ∴ 71” = 1′ 11”

and ∴ 82° 53′ 71″ = 82° 54′ 11″

(iii) 56° 38′ and 27° 42’30”

56° 38′ + 27° 42’30”

= 84° 20’ 30”           ….(1° = 60′)

56° 38′ 0”
+ 27° 42’30”
84° 80’ 30”
20′

(iv) 47° and 61° 17’4″

47° 0′ 0”

+ 61° 17’ 4”
108° 17′ 4”

#### Question -5.

In the figure, given below name :
(i) three pairs of adjacent angles.
(ii) two acute angles,
(iii) two obtuse angles
(iv) two reflex angles.

(i) Three pairs of adjacent angles are
∠AOB and ∠BOC;
∠BOC and ∠COD;
∠COD and ∠DOA.
(ii) Two acute angles are
∠AOB and ∠AOD.
(iii) Two obtuse angles are
∠BOC and ∠COD.
(iv) Two reflex angles are
∠AOB and ∠COD.

Question- 6.

In the given figure ; PQR is a straight line. If :
(i) ∠SQR = 75° ; find ∠PQS.
(ii) ∠PQS = 110°; find ∠RQS

(i) From figure,
∠PQS + ∠SQR = 180° [Linear pair of angles] ⇒∠PQS + 75° = 180°
⇒ ∠PQS = 180°-75°
⇒ ∠PQS = 105°
(ii) From figure again,
∠PQS + ∠RQS = 180°
⇒ 110° + ∠RQS = 180°
∠RQS = 180°- 110°
∠RQS = 70°

#### Question -7.

In the given figure ; AOC-is a straight line. If angle AOB = 50°, angle AOE = 90° and angle COD = 25° ; find the measure of :
(i) angle BOC
(ii) angle EOD
(iii) obtuse angle BOD
(iv) reflex angle BOD
(v) reflex angle COE.

(i) ∠AOB + ∠BOC = 180° (Linear pairs of angle)
⇒ 50° +∠BOC = 180°
⇒ ∠BOC = 180° – 50° = 130°
⇒ ∠BOC = 130°
(ii) ∠EOD + ∠COD = 90° (∵AOE = 90°)
⇒ ∠EOD + 25° = 90°
⇒ ∠EOD + 25° = 90°
⇒ ∠EOD = 90° – 25°
⇒ ∠EOD = 65°
(iii) ∠BOD = ∠BOC + COD
= 130° + 25° = 155°
(iv) Reflex ∠BOD = 360° – ∠BOD
= 360°- 155° = 205°
(v) Reflex ∠COE = 360° – ∠COE
= 360° (∠COD + ∠EOD)
= 360° – (25° + 65°)
= 360° – 90° = 270°

#### Question- 8.

In the given figure if :
(i) a = 130° ; find b.
(ii) b = 200 ; find a.
(iii) a = 5/3 right angle, find b

(i) From figure, a + b = 360°
⇒ 130° + 6 = 360
⇒ 6 = 360°-130°
⇒ b = 230°
(ii) From figure,
a + b = 360°
⇒ a + 200° = 360°
⇒ a = 360° – 200°
⇒ a = 160°
(iii) Here, a= 53 right angle
= 53 x90° = 150°
a = 150°
Here, a + b = 360°
⇒ 150° + b = 360° (∵a = 150°)
⇒ b = 360° -150°
b = 210°

#### Question -9.

In the given diagram, ABC is a straight line.
(i) If x = 53°, find y.
(ii) If y =112 right angles ; find x.

(i) From the figure,
∠ABD + ∠DBC = 180° (Linear pair of angles)
⇒ x+y=180°
⇒ 53°+y = 180° (∵ x = 53”)
⇒ y = 180° – 53°
⇒ y = 127°
(ii) From figure again,
x+y= 180°
1 + 32 x 90 = 180°
⇒ x+112 right angles = 180°
⇒ x+32 x90=180°
⇒ x + 135°= 180°
⇒ x= 180° – 135°
⇒ x = 45°

#### Question -10.

In the given figure, AOB is a straight line. Find the value ofx and also answer each of the following :
(i) ∠AOP = ……..
(ii) ∠BOP = ……..
(iii) which angle is obtuse ?
(iv) which angle is acute ?

∠AOP = x + 30°
∠BOP = x – 30°
But ∠AOP + ∠BOP = 180° (∵ ∠AOB is a straight angle)
⇒ x + 30°+x-30° = 180°
⇒ 2x = 180°
⇒ x = 90°
(i) ∠AOP = x + 30° = 90° + 30° = 120°
(ii) ∠BOP = x- 30° = 90° – 30° = 60°
(iii) ∠AOP is an obtuse angle
(iv) ∠BOP is an acute angle

#### Question- 11.

In the given figure, PQR is a straight line. Find x. Then complete the following:
(i) ∠AQB = ……..
(ii) ∠BQP = ……..
(iii) ∠AQR = …….

PQR is a straight line
∠AQP=x + 20°
∠AQB = 2x + 10°
∠BQR = x – 10°
But ∠AQP + ∠AQB + ∠BQR = 180°
⇒ x + 20° + 2x + 10° + x-10°= 180°
⇒ 4x + 20°= 180°
⇒4x= 180°-20°= 160°
⇒ x = 1604 = 40°
(i) ∠AQB = 2x + 10° = 2 x 40° + 10° = 80° + 10° = 90°
∠AQP = x + 2(T = 40° + 20° = 60°
∠BQR = x – 10° = 40° – 10° = 30°
(ii) ∠BQP = ∠AQP + ∠AQB = 60° + 90° = 150°
(iii) ∠AQR = ∠AQB + ∠BQR = 90° + 30° = 120°

#### Question- 12.

In the given figure, lines AB and CD intersect at point O.
(i) Find the value of ∠a.
(ii) Name all the pairs of vertically opposite angles.
(iii) Name all the pairs of adjacent angles.
(iv) Name all the reflex angles formed and write the measure of each.

Two lines AB and CD intersect each other at O
∠AOC = 68°
(i) ∵ AOB is a line
∠AOC + ∠BOC = 180°
⇒ 68° + a = 180°
⇒ a= 180°-68° = 112°
(ii) ∠AOC and ∠BOD and ∠BOC and ∠AOD are the two pairs of vertically opposite angles .
(iii) ∠AOC and ∠BOC; ∠BOC and ∠BOD; ∠BOD and ∠DOA;
∠DOA and AOC are the pairs of adjacent angles
(iv) ∠BOC and ∠DOA are reflex angles and also ∠AOC and ∠BOD are also reflex angles
Ref. ∠BOC = 180° + 68° = 248°
Ref. ∠DOA = 180° + 68° = 248°
Ref. ∠AOC = 180° + 112° = 292°
and ref. ∠BOD =180° + 112° = 292°

#### Question- 13.

In the given figure :
(i) If ∠AOB = 45°, ∠BOC = 30° and ∠AOD= 110°;
find : angles COD and BOD.
(ii) If ∠BOC = ∠DOC = 34° and ∠AOD = 120° ;
find : angle AOB and angle AOC.
(iii) If ∠AOB = ∠BOC = ∠COD = 38°
find : reflex angle AOC and reflex angle AOD.

(i) ∠COD = ∠AOD – ∠AOC
= ∠AOD – (∠AOB + ∠BOC)
= 110°-(45°+ 30°)
= 110°-75° = 35°
∠BOD = ∠AOD -∠AOB
= 110° -45°
= 65°
(ii) ∠AOB = ∠AOD-∠BOD
= ∠AOD – (∠BOC + ∠COD)
= 120° – (34° + 34°)
= 120°-68°
= 52°
∠AOC = ∠AOB + ∠BOC
= 52° + 34°
= 86°
(iii) Reflex ∠AOC = 360°-∠AOC
= 360° – (∠AOB + ∠BOC)
= 360° – (38° + 38°)
= 360° – 76° = 284°
Reflex ∠AOD = 360° – ∠AOD
= 360° (∠AOB + ∠BOC + ∠COD)
= 360° – (38° + 38° + 38°)
= 360°- 114°
= 246°

### ICSE Class-6th Concise Mathematics Selina Solutions Exercise – 24 BAngles (With their Types)

#### Question- 1.

Write the complement angle of :
(i) 45°
(ii) x°
(iii) (x – 10)°
(iv) 20° + y°

(i) Complement angle of 45°
= 90° – 45° = 45°
(ii) Complement angle of x°
= 90° – x° = (90 – x)°
(iii) Complement angle of (x – 10)° = 90° (x – 10°)
= 90°-x + 10° = 100°-x
(iv) Complement angle of 20° + y°
= 90°-(20°+y°)
= 90° – 20° -y° = 70° -y°

#### Question -2.

Write the supplement angle of :
(i) 49°
(ii) 111°
(iii) (x – 30)°
(iv) 20° + y°

(i) Supplement angle of 49°
= 180°-49° = 131°
(ii) Supplement angle of 111°
= 180°- 1110 = 69°
(iii) Supplement of (x – 30)° = 180° – ( x° – 30°)
= 180o – x° + 30° – 210° – x°
(iv) Supplement of ∠20° + y° = 180° – (20° +y°)
= 180° -20°-y°
= 160° -y°

#### Question- 3.

Write the complement angle of :

(i) Complement angle of 12  of 60°

= 90° – (12  of 60∘)

= 90° – 30°

= 60°

(ii) Complement angle of 15  of 160°

= 90° – (15 × 160∘)

= 90° – 32°

= 58°

(iii) Complement angle of 25  of 70°

= 90° – (25 × 70∘)

= 90° – 28°

= 62°

(iv) Complement angle of 16  of 90°

= 90° – (16 × 90∘)

= 90° – 15°

= 75°

#### Question- 4.

Write the Supplement angle of :

(i) Supplement angle of 50% of 120°

= 180° – (50% of 120°)

= 180° – (120°×50100)

= 180° – 60°

= 120°

(ii)  Supplement angle of (13 of 150∘)

= 180° – (13  of 150°)

= 180° – 50°

= 130°

(iii) Supplement angle of 60% of 100°

= 180° – (60% of 100°)

= 180° – (60100 ×100)

= 180° – 60°

= 120°

(iv) Supplement angle of 34 of 160°

= 180° – (34 of 160°)

= 180° – 120°

= 60°

Question -5.

Find the angle :
(i) that is equal to its complement ?
(ii) that is equal to its supplement ?

(i) that is equal to its complement

Let angle be x

and complements is also x

angle + its complement = 90

x + x =90

2x = 90

x = 902
x = 45

Hence 45° is equal to its complement.

(ii) that is equal to its supplement

Let angle be x

and supplement is also x

angle + its supplement = 180

x + x =180

2x = 180

x = 1802
x = 90

Hence 90° is equal to its supplement.

#### Question -6.

Two complementary angles are in the ratio 7 : 8. Find the angles.

Let two complementary angles are 7x and 8x

∴ 7x + 8x = 90°

⇒ 15x = 90°

⇒ x = 9015

⇒ x = 6°

∴ Two complementary angles are

7x = 7 x 6° = 42°

8x = 8 x 6° = 48°

#### Question -7.

Two supplementary angles are in the ratio 7 : 11. Find the angles.

Let two supplementary angles are 7x and 11x

∴ 7x+ 11x= 180°

⇒ 18x = 180°

⇒ x =18018

⇒ x = 10°

Two supplementary angles are

7x = 7 x 10° = 70°

11x= 11 x 10°= 110°

#### Question- 8.

The measures of two complementary angles are (2x – 7)° and (x + 4)°. Find x.

We know that, sum of two complementary angles = 90°

∴(2x – 7) + (x + 4) = 90°

2x – 7 + x + 4 = 90°

⇒ 2x + x – 7 + 4 = 90°

⇒ 3x – 3 = 90°

⇒ 3x = 90 + 3

⇒ 3x = 93

⇒ x = 933

x = 31

#### Question- 9.

The measures of two supplementary angles are (3x + 15)° and (2x + 5)°. Find x.

We know that, sum of two supplementary angles = 180°

∴ (3x + 15)° + (2x + 5)° = 180° ‘

3x + 15 + 2x + 5 = 180°

⇒ 3x + 2x+15 + 5 = 180°

⇒ 5x°+ 20° = 180°

⇒ 5x = 180° – 20°

⇒ 5x= 160°

⇒ x = 1605

⇒ x = 32°

#### Question- 10.

For an angle x°, find :
(i) the complementary angle
(ii) the supplementary angle
(iii) the value of x° if its supplementary angle is three times its complementary angle.

For an angle x,
(i) Complementary angle of x° = (90° – x)

(ii) Supplementary angle of x° = (180° – x)

(iii) ∵‘Supplementary angle = 3 (complementary anlge)

180° – x = 3 (90°- x)

⇒ 180°- x = 270°- 3x

⇒-x + 3x = 270°- 180°

⇒ 2x = 90°

⇒ x = 902 = 45°

∴ x = 45°

### Revision Exercise Chapter-24 Angles (With their Types) for ICSE Class-6th Concise Selina Mathematics Solutions

#### Question -1.

Explain what do you understand by :
(ii) Complementary angles ?
(iii) Supplementary angles ?

(i) Adjacent Angles:  Two angles are called adjacent angles if (a) they have a common vertex (b) they have one common arm and (iii) the other two arms of the angles are on the opposite sides of the common arm.
(ii) Complementary Angles : Two angles whose sum is 90° are called complementary angles to each other.
(iii) Supplementary Angles : Two angles whose sum.is 180° are called supplementary angles to each other.

#### Question- 2.

Find the value of ‘x’ for each of the following figures :

(i) In given figure, BOC is a straight line

∴ ∠AOC + ∠AOB = 180°

⇒ 75° + 5x + 20° = 180°

⇒ 5x + 95° = 180°

⇒ 5x = 180° – 95° = 85°

⇒ x = 855 =17∘

∴ x = 17°

(ii)

In given figure, angles are on a point

∴ The sum = 360°

⇒ 75° + 2x + 65° + 3x + x = 360°

⇒ 6x + 140° = 360°

⇒ 6x = 360° – 140° = 220°

(iii) In given figure, angles are on a point

∴ Their sum = 360°

⇒ 5x + 3x + 40°+ 120° = 360°

⇒ 8x + 160° = 360°

⇒ 8x = 360° – 160° = 200°

⇒ x = 2008=25∘

⇒ x = 25°

#### Question -3.

Find the number of degrees in an angle that is (i) $\frac { 3 }{ 5 }$ of a right angle (ii) 0.2 times of a straight line angle.

(i) 35 of a right angle

=35×90∘     (∵ 1 right angle = 90°)

= 3 × 18° = 54°

(ii) 0.2 times of a straight line angle

= 0.2 of 180°  (Straight line angle = 180°)

=210×280∘=36∘

#### Question- 4.

In the given figure; AB, CD and EF are straight lines. Name the pair of angles forming :
(i) straight line angles.
(ii) vertically opposite angles.

(i)In the given figure, AB, CD and EF are straight lines on intersecting, angles are formed a, b, c, d, l, m, n and p.

In the figure pairs of straight-line angles are ∠a, ∠b; ∠b, ∠c; ∠c, ∠d; ∠d, ∠a ∠l, ∠m; ∠m, ∠n;∠n, ∠p and ∠p, ∠l.

(ii) In the given figure, AB, CD and EF are straight lines on intersecting, angles are formed a, b, c, d, l, m, n and p.

Pairs of vertically angles are ∠a, ∠c; ∠b, ∠d; ∠l, ∠n; ∠m, ∠p.

#### Question- 5.

Find the complement of :

(i) We know that two angles are called complementary when their sum is 90°. Now

25 of 210∘

=25×210∘=2×42∘=84∘

∴ Its complement will be = 90° – 84° = 6°

(ii) We know that two angles are called complementary when their sum is 90°. Now

0.4 times of 130° = 410×130∘

= 4 × 13° = 52°

∴ Its complement will be = 90° – 52° = 38°

#### Question- 6.

Find the supplement of :

(i) We know that  two supplementary angles are those whose sum is 180°. Now

57 of 154°

=57×154°=5×22°=110°

∴ Its supplement = 180° – 110° = 70°

(ii) We know that two supplementary angles are those whose sum is 180°. Now

0.7 times of 150°

=710×150°=7×15°=105°

∴ Its supplement = 180° – 105° = 75°

#### Question -7.

Two complementary angles are in the ratio 8 : 7. Find the angles.

Let two complementary angles are 7x and 8x

∴ 7x + 8x = 90°

⇒ 15x = 90°

⇒ x = 9015

⇒ x = 6°

∴ Two complementary angles are

7x = 7 x 6° = 42°

8x = 8 x 6° = 48°

#### Question-  8.

Two supplementary angles are in the ratio 7 : 5. Find the angles.

Ratio between two supplementary angles are 7 : 5

∴ Sum of ratios = 7 + 5 = 12

But sum of two supplementary angles = 180°

∴ First angle = 180∘×712=15∘×7=105∘

and second angle = 180∘×512=15∘×5=75∘

∴ Two supplementary angles are 105° and 75°

#### Question- 9.

Two supplementary angles are (5x – 82°) and (4x + 73°). Find the value of x.

∵ Sum of two supplementary angles = 180°

∴ (5x – 82°) + (4x + 73°) = 180°

5x – 82° + 4x + 73° = 180°

9x – 9° = 180°

9x = 180° + 9° = 189°

x = 1899=21°

∴ x = 21°

#### Question -10.

Find the angle formed by the arms of a clock at:
(i) 3 O’clock
(ii) 6 O’clock
(iii) 9 O’clock
(iv) 12 O’clock

(i) 3 O’clock

We know that sum of angles at a point = 360
and on the face of a clock there are 12 marks on it.

Now at 3 O’clock, the hour hand is on 3 while minute hand is at 12

∴ Angle between them =312×360∘

= 3 × 30° = 90°

(ii) 6 O’clock

We know that sum of angles at a point = 360° and on the face of a clock there are 12 marks on it.

Now at 6 O’clock, the hour hand is on 6 while minute hand is at 12

∴ Angle between them = 612×360∘

= 6 × 30° = 90°

(iii) 9 O’clock

We know that sum of angles at a point = 360° and on the face of a clock there are 12 marks on it.

Now at 9 O’clock, the hour hand is on 9 while minute hand is at 12

∴ Angle between them  912×360∘

= 9 × 30° = 270°

(iv) 12 O’clock

We know that sum of angles at a point = 360° and on the face of a clock there are 12 marks on it.

Now at 12 O’clock, the hour hand and minute hand are both 12°

∴ Angle between them =1212×360∘=360° or 0° as both hands coincide each other.

#### Question – 11.

For an angle y°, find :
(i) its supplementary angle.
(ii) its complementary angle.
(iii) the value of y° if its supplement is four times its complement.

(i) Its supplementary angle = 180° – y

(ii) Its complementary angle = 90° – y

(iii) ∴ supplementary angle of y = 4   …(complementary angle of y)

∴ 180° – y = 4 (90° – y)

⇒ 180° – y = 360° – 4y

⇒ 4y – y = 360° – 180°

⇒ 3y = 180°

⇒ y = 1803=60∘

∴ y = 60°

#### Question- 12.

Use the adjoining figure to find :
(i) ∠BOD
(ii) ∠AOC

(i) ∠BOD

But sum of angles on the same side of a line = 180°

∴ ∠AOD + ∠COD + ∠BOC = 180°

72° + ∠COD + 64° = 180°

∠COD = 180° – (72° + 64°)

= 180° – 136° = 44°

(i) Now ∠BOD = ∠BOC + ∠COD

= 64° + 44° = 108°

(ii) ∠AOC = ∠AOD + ∠COD

= 72° + 44° = 116°

Question- 13.

Two adjacent angles forming a linear pair are in the ratio 7:5, find the angles.

In the given figure, ∠AOC and ∠BOC are adjacent angles which form a linear pair

∴ ∠AOC  ∠BOC  = 180°

But ∠AOC : ∠BOC = 7 : 5

Let ∠AOC = 7x and ∠BOC = 5x

∴ 7x + 5x = 180°

⇒ 12x = 180°

⇒ x = 18012 =15∘

∴  ∠AOC = 7x = 7 × 15 = 105°

and ∠BOC = 5x = 5 × 15 = 75°

#### Question- 14.

Find the angle that is three times its complementary angle.

Let given angle = x

Let complementary angle of a given angle is y

∴ then x + y = 90

But x = 3y

∴ 3y + y = 90°

⇒ 4y = 90°

⇒ y = 904 = 22.5°

∴ complementary = x = 90° – 22 – 5° = 67.5°

#### Question- 15.

An angle is one-thirds of a straight line angle ; find :
(i) the angle
(ii) the complement and the supplement of the angle obtained above.

straight angle = 180°

∴ given angle = 13×180∘ = 60°

(i) complement of the given angle

= 90° – 60° = 30°

(ii) Supplement of the given angle

= 180° – 60° = 120°

End of Angles ICSE Class-6th Concise Solutions :–

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