1. In Fig. 6.28, find the values of x and y and then show that AB || CD.

Solution:

A traversal intersects two lines AB and CD . Then,

x + 50˚ = 180˚

x = 180˚ – 50˚ = 130˚

y = 130˚ (Vertically opposite angles)

Therefore, ∠x = ∠y = 130˚ (alternate angles)

⸫AB||CD

2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:

Given y : z  = 3 : 7

y = 3a and z = 7a

We know, ∠PQC = ∠DQR = y

Therefore, y + z = 180° [interior angles on the same side of the traversal]

3a + 7a = 180°

10a = 180°

a = 18°

y = 3*18° = 54°

z = 7*18° = 126°

Then, x +  y = 180° [interior angles on the same side of the traversal]

x = 180° – 54° =  126°

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3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.

Solution:

Given ∠GED = 126° . Since AB||CD.
We have, ∠GED = ∠AGE [Alternate angles]

Therefore, ∠AGE= 126°

∠EGF + ∠AGE  = ∠AGF [Adjacent angles]

∠EGF = 180° – 126° = 54°

Therefore, ∠EGF = 54°

∠GFE  + ∠EFG + ∠EGF = 180°

Since AB||CD and EF⊥ CD, therefore, ∠EFG = 90°

∠GFE = 180° – ∠EFG – ∠EGF

= 180° – 90° – 54°

Therefore, ∠GFE = 36°

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4. In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS. [Hint : Draw a line parallel to ST through point R.]

Solution:

Given PQ || ST, ∠ PQR = 110° and ∠ RST = 130° . We have to find ∠QRS.

∠RST +∠NMS = 180° [interior angles on the same side of the traversal]

130° + ∠NMS = 180°

∠NMS = 180° – 130° = 50°

∠NMS = ∠QMR = 50° [vertically opposite angles]

∠PQR + ∠MQR = 180°

∠MQR = 180° – ∠PQR

= 180° – 110° = 70°

∠MQR = 70°

∠QRS + ∠MQR + ∠NMS = 180° [ sum of interior angles is equal to 180°]

∠QRS = 180° – ∠MQR – ∠NMS

= 180° – 70° – 50°

∠QRS= 60°

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5. In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.

Solution:

Given AB || CD, ∠ APQ = 50° and ∠ PRD = 127°

∠QRD = ∠QRP + ∠PRD

180°= ∠QRP + 127°

∠QRP = 180° – 127° = 53°

∠QRP = 53°

∠APQ + ∠PQC = 180°

∠PQC = 180° – 50° = 130°

∠PQC= 130°

∠CQR = ∠PQC + ∠PQR

180°  = 130° + ∠PQR

∠PQR = 50° = x

∠PQR + ∠QPR + ∠PRQ = 180°

∠QPR = 180° – 50° – 53° = 77°

y = 77°

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6. In Fig. 6.33, 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:

Construction: Draw BM⊥PQ and CN ⊥RS

Given: PQ||RS and BM||CN

Use the property of Alternate interior angles

∠2 = ∠3 ………………(1)

∠ABC = ∠1 + ∠2

But ∠1 = ∠2 therefore, ∠ABC = ∠2 + ∠2 = 2 ∠2

Similarly, ∠BCD = ∠3 + ∠4

But ∠3 = ∠4 therefore, ∠BCD = ∠3 + ∠3 = 2 ∠3

From (1),

∠ABC = ∠DCB

These are alternate angles so that AB||CD

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