# Brilliant’s Composite Mathematics Class 6 Solutions Chapter 12

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## Brilliant’s Composite Mathematics Class 6 Solutions Chapter 12 Ray and Angles

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 12, Ray and Angles. Here students can easily find step by step solutions of all the problems for Ray and Angles, Exercise 12.1, 12.2, 12.3, 12.4, 12.5, 12.6 and 12.7 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

Ray and Angles Exercise 12.1 Solution

Question no – (1)

Solution :

(i) A ray and a segment are –

In a segment we can write AB = BA But in a ray = AB ≠ BA

(ii) A ray and a line are –

Meet of two points called a line but, A ray is a part of a line that extends indefinitely in one direction from a point.

Question no – (2)

Solution :

(i) Name all the rays shown in the Fig. 12.19, whose initial point are P. Q and R respectively.

= The name of all the Rays are – PR, QB, RB

(ii) Is ray PQ different from ray PB

= No, it is not different

(iii) Is ray PA different from ray PR

= Yes, it is different from PR

(iv) Are rays QP and QR opposite

= Yes, they are opposite.

Question no – (3)

Solution :

Three examples of an angle are –

(i) Compass angle

(ii) Pain of scissor

(iii) Hour hand and minute hand of a clock.

Question no – (4)

Solution :

Name of the angles are –

(i) ∠PQR, ∠QRP and ∠RPQ

(ii) ∠ABC, ∠BCD, ∠CDA, ∠DAB

(iii) ∠AOB, ∠AOC, ∠AOD, ∠BOC, ∠BOD,

Question no – (5)

Solution :

(i) Given, In the interior of B

= P,S,T one interior ∠B

(ii) Given, Exterior of B

= Q, G are exterior of ∠B

(iii) Given, lie on B

= A, N, M, C Line on ∠B

Question no – (6)

Solution :

The arms and vertex of the following angles are –

Question no – (7)

Solution :

Correct option – (iii) on the arms

The vertex of an angle lie on its arms.

Question no – (8)

Solution :

(i) Given, 1

= ∠ABC

(ii) Given, 2

= ∠BCD

(iii) Given, 3

= ∠BAD

(iv) Given, 4

= ∠ADC

Question no – (9)

Solution :

(i) Point B is in the interior of ∠AOC → True

(ii) Point B is in the exterior of ∠AOC → False

(iii) Point C is in the interior of ∠BOD → True

(iv) Point C is in the exterior of ∠COD → False

(v) Point A is in the exterior of ∠DOB → True

(vi) Point A is in the interior of ∠AOD → False

Ray and Angles Exercise 12.2 Solution

Question no – (1)

Solution :

Figure – (i)

= ∠1 > ∠2

Figure – (ii)

= ∠3 > ∠4

Figure – (iii)

= ∠6 > ∠5

Figure – (iv)

= ∠7 > ∠8

Question no – (2)

Solution :

(a) Is 21 < 22

= No

(b) Is 21> 24

= No

(c) Is <b> <a

= Yes

(d) Is ZAOB > ZPQR

= No

(e) In which pairs the angles are equal

= In pain (i) the angles are equal.

Question no – (3)

Solution :

Required figure,

(i) One right angle

= One right angle – She look east – North

(ii) Three right angles

= There right angle – she look North – West

Question no – (4)

Solution :

(a) Two acute angles.

= ∠A, and ∠C

(b) Two obtuse

= ∠ADC, and ∠ABC

Question no – (5)

Solution :

(i) Given, 45°

= acute angle

(ii) Given, 95°

= obtuse angle

(iii) Given, 90°

= right angle

(iv) Given, 135°

= obtuse angle

(v) Given, 265°

= rectangle

(vi) Given, 101.5°

= obtuse angle

(vii) Given, 360°

= Complete angle

(viii) Given, 180°

= straight angle

(x) Given, 10°

= acute angle

(xi) Given, 27°

= acute angle

(xii) Given, 189°

= reflex angle

Question no – (6)

Solution :

(i) North and South

= Straight angle

(iiSouth and West

= right angle

(iii) South and South-East

= acute angle

(iv) North and South-East

= obtuse angle

(v) East and West.

= Straight angle

Question no – (7)

Solution :

Using only a ruler, a right angle, obtuse angle, a straight angle and an acute angle are –

Question no – (8)

Solution :

(i) 3 O’clock

= Right angle.

(ii) 1 O’clock

= Acute angle.

(iii) 12 O’clock

= eco angle.

(iv) 6 O, Clock

= Straight angle.

Required figures,

Question no – (9)

Solution :

(a) Given statement is → False.

(b) Given statement is → False.

(c) Given statement is → True.

(d) Given statement is → True.

(e) Given statement is → False.

(f) Given statement is → True.

Ray and Angles Exercise 12.4 Solution

Question no – (1)

Solution :

(i) Given, 68 °

= 90 – 68

= 22°

Complement is 22°

(ii) Given, 18 °

= 90° – 18°

= 72°

Complement is 72°

(iii) Given, 23 °

= 90° – 23°

= 67°

Complement is 67°

(iv) Given, 54 °

= 90° – 54°

= 36°

Complement is 36°

(v) Given, 65 °

= 90° – 65°

= 25°

Complement is 25°

(vi) Given, 85 °

= 90° – 85°

= 5°

Complement is 5°

Question no – (2)

Solution :

(i) Given, 95°

= 180° – 95°

= 85°

Supplement is = 85°

(ii) Given, 72°

= 180° – 72°

= 108°

Supplement is = 108°

(iii) Given, 135°

= 180° – 135°

= 45°

Supplement is = 45°

(iv) Given, 163°

= 180° – 163°

= 5°

Supplement is = 5°

(v) Given, 176°

= 180° – 176°

= 4°

Supplement is = 4°

Question no – (3)

Solution :

(i) Given, 12° , 78°

= 12° + 76°

= 90° (Complementary angle)

(ii) Given, 25°, 165°

= 25° + 165

= 180° (Supplementary angle)

(iii) Given, 30°. 150°

= 30° + 150°

= 180° (Supplementary angle)

(iv) Given, 30°, 60°

= 30° + 60°

= 90° (Complementary angle)

(v) Given, 136°, 44°

= 136° + 44°

= 180° (Supplementary angle)

(vi) Given, 55°, 35°

= 55° + 35°

= 90° (Complementary angle)

(vii) Given, 115°, 65°

= 115° + 65°

= 180° (supplementary angle)

(viii) Given, 29°, 61°

= 29° + 61°

= 90° (complementary angle)

(4) Find the angle which is :

Solution :

(i) Equal to its complement.

Let the angle be x

= x + x = 40°

= 2x = 90°

= x = 45°

Therefore, the required angle is 45°

(ii) Two times its complement.

Let angle x

= 90° – x = 2x

= 2x + x = 90°

= 3x = 90°

= x = 30°

Two times its complement angle,

= 2 × 30

= 60°

(iii) 2/3 of its complement.

2/3 + x = 90°

= 2x + 3 = 90 × 3

= 90° × 3/5

= 40

Required Angle,

= 2/3 × 48

= 36°

Thus, the angle is 36°

Question no – (5)

Solution :

(i) Equal to supplement,

= x + x = 180°

= 2x = 180°

= x = 90°

The required angle is 90°

(ii) Three times its supplement,

= 9x + x = 180°

= 4x = 180°

= x = 45°

The required angles,

= 3 × 45

= 135°

(iii) Five times its supplement,

= 5x + x = 180°

= 6x = 180°

= x = 180/6

= x = 30°

The required angle,

= 5 × 30

= 150°

Question no – (6)

Solution :

Fig – 12.49 (i)

In the given figure three are No adjacent angles.

Question no – (7)

Solution :

Fig – 12.49 (ii)

= ∠AOC and ∠AOB are not adjacent angles because There are no common arm.

Question no – (8)

Solution :

(i) acute angle

= Acute angle – 45°

From a linear pair

Other angles,

= 180° – 45°

= 135° obtuse angle

(ii) obtuse angle

= obtuse angle – 110°

Other angles from a linear pair,

= 180° – 110°

= 70° acute angle

(iii) right angle.

= right angle = 90°

Other angle from a linear pair,

= 180° – 90°

= 90° right angle

Question no – (9)

Solution :

Let the angles are = x, 2x

= x + 2x = 180°

= 3x = 180°

= x = 60°

Angles are = 60, 120°

Question no – (10)

Solution :

Other angles are –

= ∠3 + ∠2 = 180°

= 25 + ∠2 = 180°

= ∠2 = 180 – 25°

= 135°

∠3 = ∠1 = 25°

∠2 = ∠54 = 135°

Question no – (11)

Solution :

(i) Vertical angles,

= ∠1 and ∠3, ∠2 and ∠4

= ∠5 and ∠7, ∠6 and ∠8

(ii) Vertical angles,

= ∠1 and ∠4, ∠7 and ∠8

= ∠3 and ∠6, ∠11 and ∠12

= ∠2 and ∠5, ∠9 and ∠10

Question no – (12)

Solution :

(i) Each pair of vertical opposite angles,

∠1 and ∠4, ∠2 and ∠3, ∠5 and ∠7, ∠6 and 8 are vertical opposite

(ii) Each linear pair,

∠1 and ∠2, ∠3 and ∠4, ∠5 and ∠6, ∠7 and ∠8

∠1 and ∠3, ∠2 and ∠4, ∠5 ad ∠8, ∠7 and 6 are linear pairs.

(iii) Set of angles at a point,

∠1, ∠2, ∠3 and ∠4; ∠5, ∠6, ∠7 and ∠8

Question no – (13)

Solution :

Figure – (i)

= x° + 45° = 180°

= x = 180° – 45°

= 135°

Figure – (ii)

= x° + 75° + 50° = 180°

= x + 125° = 180°

= 180° – 125°

= 55°

Figure – (iii)

= x° + 70° + 65° + 110° = 360°

= x + 245° = 360°

= x = 360° – 245°

= 115°

Figure – (iv)

= 4x + 3x + 5x = 180°

= 12x = 180°

= x = 180/12

= 15°

Figure – (v)

= 2x + 15 + 3x + 25 = 180°

= 5x + 40 = 180°

= 5x = 180 – 40

= 140/5

= 28°

Figure – (vi)

= 2x + 7x + 45° = 180°

= 9x = 180 – 4

= x = 185/9

= 15°

Question no – (14)

Solution :

= Let the angle – 46° complement – 90° – 46° = 44°

= No the complements is not great then 45°

Question no – (15)

Solution :

(i) Two acute angles can be supplementary → False

(ii) Two obtuse angles can be supplementary → False

(iii) Two obtuse angles can be supplementary → False

(iv) Two angles are complementary if their sum is 180° → False

(v) Supplementary angles always form a linear pair → False

(vi) Angles of a linear pairs are supplementary → True

(vii) Angles of a linear pairs are complementary → False

(viii) Adjacent angles can be complementary → True

(ix) If two lines intersect, then there are two pairs of vertically opposite angles → True

(x) Two adjacent angles are always supplementary → False

(xi) An angle is less than 90°, its supplement will also be less than 90° → False

Next Chapter Solution :

Updated: June 12, 2023 — 6:46 am