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NCERT Solutions Ganita Prakash (Part 1) Chapter 5 –122Section 5.8 Alternate Angles — Examples

Book page 120 Updated on2026-09-05

Q1.
Example 1: In Fig. 5.26, parallel lines l and m are intersected by the transversal t. If ∠6 is 135°, what are the measures of the other angles?
Answer

Start from ∠6 = 135° and travel through the figure.

∠2 = ∠6 = 135° (corresponding angles, l ∥ m)
∠8 = ∠6 = 135° (vertically opposite)
∠4 = ∠8 = 135° (corresponding)
∠2 = ∠4 = 135° (vertically opposite) ✔

∠5 + ∠6 = 180° (linear pair)
∠5 = 180° – 135° = 45°
Similarly ∠1 = ∠3 = ∠7 = 45°
AngleMeasure
∠2, ∠4, ∠6, ∠8135°
∠1, ∠3, ∠5, ∠745°
Why it happens: Only two sizes can appear when a transversal crosses parallel lines, and they must add to 180°. Here they are 135° and 45°, and 135 + 45 = 180 ✔
Q2.
Example 2: In Fig. 5.27, lines l and m are intersected by the transversal t. If ∠a is 120° and ∠f is 70°, are lines l and m parallel to each other?
Answer

No, l and m are not parallel.

∠a + ∠b = 180° (linear pair)
120° + ∠b = 180°
∠b = 60°

∠b and ∠f are corresponding angles
For l ∥ m we would need ∠b = ∠f
But 60° ≠ 70° ✘

The corresponding angles are unequal, so the lines are not parallel.

Why it happens: ∠f is 10° larger than ∠b, which means line m leans 10° more than line l. Lines leaning differently must meet somewhere — here they would meet on the side where the gap closes.
Tip: Never compare 120° with 70° directly. First bring both angles into the same position at the two crossings, then compare.
Q3.
Example 3: In Fig. 5.28, parallel lines l and m are intersected by the transversal t. If ∠3 is 50°, what is the measure of ∠6?
Answer

∠6 = 130°.

∠2 + ∠3 = 180° (linear pair)
∠2 = 180° – 50° = 130°
∠2 = ∠6 (corresponding angles, l ∥ m)
So ∠6 = 130°

Notice the by-product: ∠3 + ∠6 = 50° + 130° = 180°. Angles ∠3 and ∠6 are called interior angles on the same side of the transversal.

Why it happens: ∠3 and ∠2 fill a straight line, and ∠2 is simply ∠6 copied across to the other parallel line. So ∠3 and ∠6 together fill exactly the same straight angle of 180°.
Q4.
Is there a relation between ∠3 and ∠6? You could try to find the relationship by taking different values for ∠3 and see what ∠6 is. Once you find a relation, try to justify it or prove that this relation holds always.
Answer

Yes: ∠3 + ∠6 = 180° always. Interior angles on the same side of the transversal are supplementary.

∠3∠2 = 180° – ∠3∠6 = ∠2∠3 + ∠6
50°130°130°180°
70°110°110°180°
90°90°90°180°
115°65°65°180°

Proof (no measurement used):

∠2 + ∠3 = 180° (linear pair on line l)
∠2 = ∠6 (corresponding angles, l ∥ m)
Replace ∠2 by ∠6 in the first line:
∠6 + ∠3 = 180°
Why it happens: The whole argument is one substitution. ∠6 is just ∠2 wearing a different name, and ∠2 was already the partner of ∠3 in a straight angle. So the sum can never be anything but 180°.
Did you know? These are also called co-interior or allied angles. Together with corresponding angles and alternate angles, they are the three facts you use to solve almost every question in this chapter.
Q5.
Example 4: In Fig. 5.29, line segment AB is parallel to CD and AD is parallel to BC. ∠DAC is 65° and ∠ADC is 60°. What are the measures of angles ∠CAB, ∠ABC, and ∠BCD?
Answer

Use the interior-angles rule twice.

Step 1 — take AB ∥ CD with transversal AD
∠ADC + ∠DAB = 180° (interior angles, same side)
60° + ∠DAB = 180°
∠DAB = 120°

Step 2 — split ∠DAB
∠DAB = ∠DAC + ∠CAB
120° = 65° + ∠CAB
∠CAB = 55°

Step 3 — take AD ∥ BC with transversal CD
∠ADC + ∠BCD = 180°
60° + ∠BCD = 180°
∠BCD = 120°

Step 4 — take AB ∥ CD with transversal BC
∠BCD + ∠ABC = 180°
120° + ∠ABC = 180°
∠ABC = 60°

So ∠CAB = 55°, ∠ABC = 60°, ∠BCD = 120°.

Why it happens: ABCD has both pairs of opposite sides parallel, so it is a parallelogram. That is why the opposite angles come out equal (∠ADC = ∠ABC = 60° and ∠DAB = ∠BCD = 120°) and why the angles next to each other add to 180°.
Check it yourself: 60 + 120 + 60 + 120 = 360°, the correct angle sum for any four-sided figure ✔
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