It is given that PQ || ST,
Draw a line XY || ST,
So, XY || PQ, i.e, PQ || ST || XY
Since, PQ || XY & QR is the transversal.
We know that,
Sum of co-interior angles = 180°.
⇒ ∠PQR + ∠QRX = 180°
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 180° - 110°
⇒ ∠QRX = 70°.
Also, ST || XY & SR is transversal
⇒ ∠SRY + ∠RST = 180° (Interior angles on the same side of the transversal are supplementary)
⇒ ∠SRY + 130° = 180°
⇒ ∠SRY = 180° - 130°
⇒ ∠SRY = 50°.
From figure,
⇒ ∠QRX + ∠QRS + ∠SRY = 180° (Linear pairs)
⇒ 70° + ∠QRS + 50° = 180°
⇒ 120° + ∠QRS = 180°
⇒ ∠QRS = 180° - 120°
⇒ ∠QRS = 60°
Hence, ∠QRS = 60°.
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BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
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Mathematics | Lines and AnglesWeb Content • Interactive Notes
Lines and Angles — Interactive Study Guide
Angle Pair Cheat Sheet
| Pair | Relationship | Condition |
|---|---|---|
| Complementary | Sum = 90° | Any two angles |
| Supplementary | Sum = 180° | Any two angles |
| Linear Pair | Sum = 180° | Adjacent + on a line |
| Vertically Opposite | Equal | Intersecting lines |
| Corresponding | Equal | Parallel lines + transversal |
| Alternate Interior | Equal | Parallel lines + transversal |
| Co-interior | Sum = 180° | Parallel lines + transversal |
Triangle Angle Facts
∠A + ∠B + ∠C = 180° (angle sum property)
Exterior angle = sum of remote interior angles
Quick Self-Check
- Two parallel lines cut by a transversal: one angle is 72°. Find all 8 angles. (72°, 108° alternating)
- In ΔABC, ∠A = 45°, ∠B = 65°. Find ∠C. (70°)
- An exterior angle of a triangle is 130°. One non-adjacent interior angle is 50°. Find the other. (80°)