# Angles ICSE Class-6th Concise Selina Maths Solutions Ch-24

## Chapter-24 Angles (With their Types) Selina Publications Solved questions

**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 CISCE 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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### Exercise – 24 A **Angles **(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.

**Answer-1**

**(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.

**Answer-2**

**(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.

**Answer-3**

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

**Question- 4.**

Add :

(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″

**Answer-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.

**Answer-5**

(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

**Answer-6**

**(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.

**Answer-7**

**(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

**Answer-8**

**(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= ^{5}⁄_{3} right angle

= ^{5}⁄_{3} 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 =1^{1}⁄_{2} right angles ; find x.

**Answer-9**

**(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 + ^{3}⁄_{2} x 90 = 180°

⇒ x+1^{1}⁄_{2} right angles = 180°

⇒ x+^{3}⁄_{2} 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 ?

**Answer-10**

∠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 = …….

**Answer-11**

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 = ^{160}⁄_{4} = 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.

**Answer-12**

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.

**Answer-13**

**(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 B **Angles **(With their Types)

**Question- 1.**

Write the complement angle of :

(i) 45°

(ii) x°

(iii) (x – 10)°

(iv) 20° + y°

**Answer-1**

**(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°

**Answer-2**

**(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 :

**Answer-3**

**(i) Complement angle of ^{1}⁄_{2} of 60°**

= 90° – (^{1}⁄_{2} of 60∘)

= 90° – 30°

= 60°

**(ii) Complement angle of ^{1}⁄_{5} of 160°**

= 90° – (^{1}⁄_{5} × 160∘)

= 90° – 32°

= 58°

**(iii) Complement angle of ^{2}⁄_{5} of 70°**

= 90° – (^{2}⁄_{5} × 70∘)

= 90° – 28°

= 62°

**(iv) Complement angle of ^{1}⁄_{6} of 90°**

= 90° – (^{1}⁄_{6} × 90∘)

= 90° – 15°

= 75°

**Question- 4.**

**Write the Supplement angle of :**

**Answer-4**

**(i) Supplement angle of 50% of 120°**

= 180° – (50% of 120°)

= 180° – (120°×^{50}⁄_{100})

= 180° – 60°

= 120°

**(ii) Supplement angle of ( ^{1}⁄_{3} of 150∘)**

= 180° – (^{1}⁄_{3} of 150°)

= 180° – 50°

= 130°

**(iii) ****Supplement angle of 60% of 100°**

= 180° – (60% of 100°)

= 180° – (** ^{60}⁄_{100 }**×100)

= 180° – 60°

= 120°

**(iv) Supplement angle of ^{3}⁄_{4} of 160°**

= 180° – (** ^{3}⁄_{4}** of 160°)

= 180° – 120°

= 60°

**Question -5.**

Find the angle :

(i) that is equal to its complement ?

(ii) that is equal to its supplement ?

**Answer-5**

(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 = ^{90}⁄_{2}

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 = ^{180}⁄_{2}

x = 90

Hence 90° is equal to its supplement.

**Question -6.**

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

**Answer-6**

Let two complementary angles are 7x and 8x

∴ 7x + 8x = 90°

⇒ 15x = 90°

⇒ x = ^{90}⁄_{15}

⇒ 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.

**Answer-7**

Let two supplementary angles are 7x and 11x

∴ 7x+ 11x= 180°

⇒ 18x = 180°

⇒ x =^{180}⁄_{18}

⇒ 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.

**Answer-8**

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 = ^{93}⁄_{3}

x = 31

**Question- 9.**

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

**Answer-9**

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 = ^{160}⁄_{5}

⇒ 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.

**Answer-10**

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 = ** ^{90}⁄_{2}** = 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 :

(i) Adjacent angles ?

(ii) Complementary angles ?

(iii) Supplementary angles ?

**Answer-1**

**(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 :

**Answer-2:**

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

∴ ∠AOC + ∠AOB = 180°

⇒ 75° + 5x + 20° = 180°

⇒ 5x + 95° = 180°

⇒ 5x = 180° – 95° = 85°

⇒ x = ** ^{85}⁄_{5 }**=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 = ** ^{200}⁄_{8}**=25∘

⇒ x = 25°

**Question -3.**

Find the number of degrees in an angle that is (i) of a right angle (ii) 0.2 times of a straight line angle.

**Answer-3**

(i) ** ^{3}⁄_{5}** of a right angle

=** ^{3}⁄_{5}**×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°)

=** ^{2}⁄_{10}**×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.

**Answer-4**

(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 :

**Answer- 5**

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

** ^{2}⁄_{5}** of 210∘

=** ^{2}⁄_{5}**×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° = ** ^{4}⁄_{10}**×130∘

= 4 × 13° = 52°

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

**Question- 6.**

Find the supplement of :

**Answer-6**

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

** ^{5}⁄_{7}** of 154°

=** ^{5}⁄_{7}**×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°

=** ^{7}⁄_{10}**×150°=7×15°=105°

∴ Its supplement = 180° – 105° = 75°

**Question -7.**

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

**Answer-7**

Let two complementary angles are 7x and 8x

∴ 7x + 8x = 90°

⇒ 15x = 90°

⇒ x = ^{90}⁄_{15}

⇒ 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.

**Answer-8**

Ratio between two supplementary angles are 7 : 5

∴ Sum of ratios = 7 + 5 = 12

But sum of two supplementary angles = 180°

∴ First angle = 180∘×** ^{7}⁄_{12}**=15∘×7=105∘

and second angle = 180∘×** ^{5}⁄_{12}**=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.

**Answer-9**

∵ Sum of two supplementary angles = 180°

∴ (5x – 82°) + (4x + 73°) = 180°

5x – 82° + 4x + 73° = 180°

9x – 9° = 180°

9x = 180° + 9° = 189°

x = ** ^{189}⁄_{9}**=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

**Answer-10**

**(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 =** ^{3}⁄_{12}**×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 = ** ^{6}⁄_{12}**×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 ** ^{9}⁄_{12}**×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 =** ^{12}⁄_{12}**×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.

**Answer-11**

(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 = ** ^{180}⁄_{3}**=60∘

∴ y = 60°

**Question- 12.**

Use the adjoining figure to find :

(i) ∠BOD

(ii) ∠AOC

**Answer-12**

(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.

**Answer-13**

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 = ** ^{180}⁄_{12 }**=15∘

∴ ∠AOC = 7x = 7 × 15 = 105°

and ∠BOC = 5x = 5 × 15 = 75°

**Question- 14.**

Find the angle that is three times its complementary angle.

**Answer-14**

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 = ** ^{90}⁄_{4}** = 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.

**Answer-15**

straight angle = 180°

∴ given angle = ** ^{1}⁄_{3}**×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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