# Number Line Prime Class 7 Solutions Chapter 15

## Number Line Prime Class 7 Solutions Chapter 15 Lines and Angles

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 15, Lines and Angles. Here students can easily find step by step solutions of all the problems for Lines and Angles, Exercise 15A and 15B 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. Here in this post students will get chapter 15 solutions. Here in this post all the solutions are based on latest Syllabus.

Lines and Angles Exercise 15A Solution :

Question no – (1)

Solution :

Complement angles Sum always 90°

(a) Given, 50°

∴ 90° – 50°

= 40° [complement angle]

(b) Given, 48°30’

∴ 90° 00’ – 48°30’

= 41°60’

(c) Given, 15°

∴ 90° – 15°

= 75°

(d) Given, 65°

∴ 90° – 65°

= 25°

(e) Given, 35°

∴ 90°- 35°

= 55°

(f) Given, 18°

∴ 90° – 18°

= 72°

Question no – (2)

Solution :

Supplement angles sum is allows 180°

(a) As per the question, 45°

∴ 180° – 45°

= 163°

(b) As Per the question, 18°

∴ 180° – 18°

= 162°

(c) As per the question, 51°

∴ 180° – 51°

= 129°

(d) Given, 65°

∴ 180° – 121°

= 59°

(e) As per the question, 176°

∴ 180° – 176°

= 4°

(f) As per the question ,91°

∴ 180° – 91°

= 89°

Question no – (3)

Solution :

Let, the angle is = x

Another angle is = 2/3 x

∴ x + 2/3 x = 180 °

∴ 5x/3 = 180

∴ x = 108°

∴ Another angle is,

= 2/3 × 108

= 72°

Question no – (5)

Solution :

Let the complementary angle = x

∴ The angle = 2x + 15°

∴ 2x + 15° + x = 90°

∴ 3x = 90° – 15

∴ x = 75/3

∴ x = 25

Required the angle

= 50 + 15

= 65°

Hence, the angles will be 65°

Question no – (6)

Solution :

Let the supplementary angle = 65°

∴ The angle = 3x + 20°

∴ 3x + 20° + x = 180°

∴ 4x = 180 – 20

∴ x = 160/5

∴ x = 40°

∴ The angle,

= 120 + 20

= 140°

Hence, the required angle will be 140°

Question no – (7)

Solution :

(a) 50 + 4x + 6x + 3x = 180°

∴ 13x = 180 – 50

∴ x = 130°/13

∴ x = 10°

∴ Angles are – 50°, 60°, 40°, 30°

(b) 38 + x + x + x + x + 50 = 180°

∴ 4x = 180° – 88

∴ x = 92/4

∴ x = 23°

∴ Angles are = 38°, 23°, 23°, 23°, 23°, 50°

Question no – (8)

Solution :

∴ 5x + 150 + 40° + 6x – 10° + 2x – 10° = 360°

∴ 13x + 35 = 360°

∴ 13x = 360 – 35

∴ x = 325/13

∴ x = 25

∴ Angles are 140° 40°, 140°, 40°

Lines and Angles Exercise 15B Solution :

Question no – (1)

Solution :

(a) ∴ y = 180° – 45°

∴ y = 135° (alternate exterior angles)

(b) ∴ y = 60° (alternate exterior angles)

∴ x = 180° – 60°

∴ x = 120°

(c) ∴ x = 180° – 65°

∴ x = 115°

∴ x = y = 115° (vertically opposite angles)

Question no – (2)

Solution :

∴ ∠MPB + ∠BPQ = 180°

Given, ∠BPQ = ∠CQP [alternate exterior angles]

= 4x + 150°

∴ 6x + 35° + 4x + 15° = 180°

∴ 10x + 50 = 180°

∴ 10x = 130

∴ x = 13°

∴ LMPB = 78 + 35

= 113°

(a) ∠APQ = ∠APQ = ∠MPB = 113° [Vertically opposite angles]

(b) ∠PQD = ∠PQD = ∠MPB = 113° [corresponding angles]

(c) ∠PQC = ∠PQC

= 52 + 15

= 67°

Question no – (3)

Solution :

Let, ∠MPB = 5x, ∠PQC = 7x

∴ ∠PQC = ∠MPQ = 7x [corresponding angle]

∴ 5x + 7x = 180°

∴ 12x = 180°

∴ x = 180°

∴ ∠MPB = 75°, ∠PQC = m ∠MPA = 105°

(a) ∠PQD = 2 MPB = 75° [Corresponding Angle]

(b) ∠MPA = 150°

Question no – (4)

Solution :

Let, ∠LQD = x  = ∠MPB [corresponding Angles]

∴ ∠MPA = 2x + 24°

∴ ∠MPA + ∠MPB = 180°

∴ 2x + 24° + x = 180°

∴ 3x = 156

∴ x = 52

(a) ∠APQ = ∠MPB = 52° [vertically opposite angle]

(b) ∠DQN = ∠MQC = ∠MPA = 2 × 52 + 24° [vertically opposite]

= 104 + 24°

= 128°

Question no – (5)

Solution :

∴ Here ∠OMN = ∠MOD = 20° [vertically opposite angle]

∴ ∠MOC = 180° – 20°

∴ ∠MOC = 160°

And, ∠BOC + ∠MOC = 310°

∴ ∠BOC = 310° – 160°

= 150°

∴ ∠OBA = 180° – 15° = 30° [Corresponding Angles]

(A) ∠OBA = 30°

(B) ∠BOC = 150°

Next Chapter Solution :

Updated: July 25, 2023 — 6:12 am