RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Question 1.
In the figure, find the values of x and y and then show that AB ∥ CD.

Answer:
x and 50° form a linear pair,
∴ x + 50° = 180°
⇒ x = 180° – 50° = 130°………………. (1)
Also, y = 130° (Vertically opposite angles) ……. (2)
∴x = y [From (1) and (2)]
But x and y are alternate angles.
Therefore, AB ∥CD.
Hence Proved.

Question 2.
In the figure, if AB ∥ CD, CD ∥ EF and y : z = 3:7, find x.

Answer:
Since CD ∥ EF and transversal PQ intersects them at S and T respectively, therefore
∠CST = ∠STF . (Alternate angles)
⇒ 180° – y = z
(∵ ∠y + ∠CST = 180° being linear pair)
⇒ y + z = 180

Given y: z = 3:7.
Let y = 3a, z = 7a
3a + 7a = 180°
10a = 180° A B
⇒ a = 18°
y = 3 × 18 = 54°,
z = 7 × 18= 126°
y = 54°, z = 126°

Since AD ∥ CD and transversal PQ intersects them at R and S respectively, therefore
∠ARS + ∠RSC = 180° (Interior angles on the same side of the transversal are supplementary.)
or x + y = 180°
⇒ x = 180° – y = 180° – 54° = 126° (∵ y = 54°)
Hence, x = 126°.

Question 3.
In the figure, if AB ∥ CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Answer:
Here, AB ∥ CD and transversal GE cuts them at G and E respectively.
∠AGE ~ ∠GED (Alternate angles)
⇒ ∠AGE = 126° [∵ ∠GED = 126° (given)]
∠GEF = ∠GED – ∠FED = 126° – 90° = 36° C
and, ∠FGE = ∠GEC (Alternate angles)
⇒ ∠FGE – 90° – ∠GEF
= 90°- 36° = 54°
Hence, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°.

Question 4.
In the figure, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
(Hint: Draw a line parallel to ST through point R.)

Answer:
Produce PQ to intersect ST in a point M. Now, PM ∥ ST and transversal SM intersects them at M and S respectively.
∠SMQ = ∠TSM (Alternate angles)
⇒ ∠SMQ = 130°
⇒ ∠QMR = 180° – 130° = 50° (∵ ∠SMQ + ∠QMR = 180°, linear pair)

Now, ray QR stands at Q on PM.
∠PQR + ∠RQM = 180°
⇒ 110° + ∠RQM = 180°
⇒ ∠RQM = 70°
∠QRS = 180° – (70° + 50°) = 60° (∵ Sum of the angles of a triangle is 180°)

Question 5.
In the figure, if AB ∥ CD, ∠APQ = 50° ∠PRD = 127°, find x and y.

Answer:
Here, AB ∥ CD and transversal PQ intersects them at P and Q respectively.
∴ ∠PQR = ∠APQ (Alternate angles)
⇒ x = 50° [∵ ∠APQ = 50° (given)]

Also, AB ∥ CD and transversal PR intersects them at P and R respectively.
∴ ∠APR = ∠PRD (Alternate angles)
⇒ ∠APQ + ∠QPR = 127° (∵ ∠PRD = 127°)
⇒ 50° + y = 127° (∵∠APQ = 50°)
⇒ y = 127° – 50° = 77°
Hence, x = 50° and y = 77°.

Question 6.
In the 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.

Answer:
Two plane mirrors PQ andBS are placed parallel to each other, i.e. PQ paths BC and CD.
Let BN and CM are the normals to the plane mirrors PQ and RS respectively. Since BN ⊥ PQ, CM ⊥ RS and PQ ∥ RS, therefore BN ⊥ RS ⇒ BN ∥ CM
Thus, BN and CM are two parallel lines and a transversal BC cuts them at B and C respectively.
∴ ∠2 = ∠3 (Alternate interior angles)

But, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)
∠1 + ∠2 – ∠2 + ∠2
and ∠3 + ∠4 – ∠3 + ∠3
⇒ ∠1 + ∠2 = 2(∠2)
and ∠3 + ∠4 = 2(∠3) ‘
⇒ ∠1 + ∠2 = ∠3 + ∠4
∠ABC = ∠BCD
Thus, lines AB and CD are intersected by transversal BC, such that ∠ABC = ∠BCD.
i.e., alternate interior angles are equal.
Therefore, AB ∥ CD.