**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 – Lines and Angles**

**Q.1 In given figure, find the values of x and y and then show that AB || CD.**

**Solution:**

x + 50˚ = 180˚ [ Linear Pair of Angles ]

x = 180˚ – 50˚

x = 130˚

y = 130˚ [ Vertically Opposite Angles ]

**x = y = 130****˚
**x and y are Alternate interior angle for Line AB and CD

**AB || CD**

**Q.2 In given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.**

**Given :**

AB || CD

CD || EF

i.e, AB || CD || EF

y : z = 3 : 7

**To Find : **x = ?

**Solution:**

x = z [ Alternate Interior Angles ]

y : z = 3 : 7

Let, y = 3a and z = 7a

x + y = 180˚ [ Co interior Angles ]

z + y = 180˚ [ x = z from above]

7a + 3a = 180˚

10 a = 180˚

a = 180˚/10

a = 18˚

x = z = 7a = 7 x 18˚ = 126˚

**x =126****˚**

**Q.3 In given figure, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE**

**Given :**

AB || CD

EF ⊥ CD

∠ GED = 126°

**To Find :**

∠ AGE = ?

∠ GEF = ?

∠ FGE = ?

**Solution:**

**∠ AGE = ∠ GED = 126° **[ Alternate Interior Angles ]

∠ GED = 126° [ Given ]

∠ GEF + ∠ DEF = 126°

∠ GEF + 90° = 126°

∠ GEF = 126° – 90°

**∠ GEF = 36°
**∠ AGE + ∠ FGE = 180° [ Linear Pair of Angles ]

126° + ∠ FGE = 180°

∠ FGE = 180° – 126°

**∠ FGE = 54°**

**Q. 4 In given figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.**

**Given :**

PQ || ST

∠ PQR = 110°

∠ RST = 130°

**To Find : **∠ QRS = ?

**Solution:**

Let us draw XY || ST passing through point R

∠ PQR + ∠ XRQ = 180° [ Co interior Angles ]

110° + ∠ XRQ = 180°

∠ XRQ = 180° – 110° = 70°

∠ XRQ + ∠ QRS = 130° [ Co interior Angles ]

70° + ∠ QRS = 130°

QRS = 130° – 70°

**QRS = 60°**

**Q.5 In given figure, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.**

**Given :**

AB || CD

∠ APQ = 50°

∠ PRD = 127°

**To Find :**

x = ?

y = ?

**Solution:**

∠ APQ = ∠ PQR = 50° [ Alternate interior Angles]

**x = 50°
**∠APR = ∠ PRD = 127° [ Alternate interior Angles]

∠APQ + ∠ QPR = 127°

50° + ∠ QPR = 127°

∠ QPR = 127° – 50°

**∠ QPR = 77°**

**Q.6 In given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD**

**Solution:**

Let us draw, BM ⊥ PQ, CN ⊥ RS

PQ || RS [ Given ]

Therefore, BM || CN

∠ 2 = ∠ 3 [ Alternate interior Angles]

∠ 1 = ∠ 2, ∠ 3 = ∠ 4 [By Laws of Reflection]

∠1 = ∠ 2 = ∠3 = ∠ 4

Also, ∠1 + ∠ 2 = ∠3 + ∠ 4

∠ ABC = ∠ CDB

∠ ABC & ∠ CDB are Alternate interior angles

**AB || CD**

**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 – Lines and Angles**