NCERT Solutions for Class 7th Maths Chapter 5 Examples — Alternate Angles
Book page 120–122 Updated on2026-09-19
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?
Fig. 5.26, page 121 — parallel lines l and m cut by transversal t, with ∠6 = 135°.
Answer
Start from ∠6 = 135° and travel through the figure.
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?
Fig. 5.27, page 121 — lines l and m cut by transversal t, with ∠a = 120° and ∠f = 70°.
∠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?
Fig. 5.28, page 121 — parallel lines l and m cut by transversal t, with ∠3 = 50°.
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?
Fig. 5.29, page 122 — AB ∥ CD and AD ∥ BC, with diagonal AC; ∠DAC = 65° and ∠ADC = 60°.
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 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 ✔