Looking for accurate SEBA Class 9 Maths Chapter 6 Exercise 6.3 (Lines and Angles) textual solutions for the 2026–27 academic year? Access step-by-step solutions for the angle sum property of a triangle, exterior angle theorem, and step-by-step geometric proofs designed to help you score top marks in your annual exams.
EXERCISE 6.3
1. In Fig. 6.35, sides QP and RQ of △PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Given:
∠SPR = 135°
∠PQT = 110°
Solution:
∠QPR + ∠SPR = 180° (Linear pair)
∠QPR + 135° = 180°
∠QPR = 180° – 135° = 45°
∠PQR + ∠PQT = 180° (Linear pair)
∠PQR + 110° = 180°
∠PQR = 180° – 110° = 70°
In △PQR,
∠QPR + ∠PQR + ∠PRQ = 180° (Angle sum property)
45° + 70° + ∠PRQ = 180°
115° + ∠PRQ = 180°
∠PRQ = 180° – 115° = 65°
Answer: ∠PRQ = 65°
SEBA Class 9 Maths Chapter 6.3 Solutions
2. In Fig. 6.36, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of △XYZ, find ∠OZY and ∠YOZ.
Given:
∠X = 62°
∠XYZ = 54°
YO and ZO are bisectors of ∠XYZ and ∠XZY respectively.
Solution:
In △XYZ,
∠X + ∠XYZ + ∠XZY = 180° (Angle sum property)
62° + 54° + ∠XZY = 180°
116° + ∠XZY = 180°
∠XZY = 180° – 116° = 64°
Since ZO bisects ∠XZY,
∠OZY = 1/2 ∠XZY = 1/2 × 64° = 32°
Since YO bisects ∠XYZ,
∠OYZ = 1/2 ∠XYZ = 1/2 × 54° = 27°
In △YOZ,
∠OYZ + ∠OZY + ∠YOZ = 180° (Angle sum property)
27° + 32° + ∠YOZ = 180°
59° + ∠YOZ = 180°
∠YOZ = 180° – 59° = 121°
Answer: ∠OZY = 32° and ∠YOZ = 121°
Class 9 Maths Chapter 6 Exercise 6.3 Question Answer SEBA 2026-27
3. In Fig. 6.37, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Given:
AB || DE
∠BAC = 35°
∠CDE = 53°
Solution:
Since AB || DE and AE is a transversal,
∠CED = ∠BAC (Alternate interior angles)
∠CED = 35°
In △CDE,
∠CDE + ∠CED + ∠DCE = 180° (Angle sum property)
53° + 35° + ∠DCE = 180°
88° + ∠DCE = 180°
∠DCE = 180° – 88° = 92°
Answer: ∠DCE = 92°
Lines and Angles Class 9 SEBA Ex 6.3 Solutions
4. In Fig. 6.38, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Given:
∠PRT = 40°
∠RPT = 95°
∠TSQ = 75°
Solution:
In △PRT,
∠RPT + ∠PRT + ∠PTR = 180° (Angle sum property)
95° + 40° + ∠PTR = 180°
135° + ∠PTR = 180°
∠PTR = 180° – 135° = 45°
∠STQ = ∠PTR = 45° (Vertically opposite angles)
In △TSQ,
∠TSQ + ∠STQ + ∠SQT = 180° (Angle sum property)
75° + 45° + ∠SQT = 180°
120° + ∠SQT = 180°
∠SQT = 180° – 120° = 60°
Answer: ∠SQT = 60°
Angle Sum Property Class 9 SEBA Ex 6.3 Notes
5. In Fig. 6.39, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Given:
PQ ⊥ PS ⟹ ∠QPS = 90°
PQ || SR
∠SQR = 28°
∠QRT = 65°
Solution:
Since PQ || SR and QR is a transversal,
∠PQR = ∠QRT (Alternate interior angles)
x + 28° = 65°
x = 65° – 28° = 37°
In △PQS,
∠QPS + x + y = 180° (Angle sum property)
90° + 37° + y = 180°
127° + y = 180°
y = 180° – 127° = 53°
Answer: x = 37°, y = 53°
Class 9 Maths Chapter 6 Exercise 6.3 Answers
6. In Fig. 6.40, the side QR of △PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.
Given:
QT is the bisector of ∠PQR ⟹ ∠PQR = 2∠TQR
RT is the bisector of ∠PRS ⟹ ∠PRS = 2∠TRS
To Prove:
∠QTR = 1/2 ∠QPR
Proof:
In △QTR, ∠TRS is an exterior angle.
∠TRS = ∠QTR + ∠TQR (Exterior angle property)
∠QTR = ∠TRS – ∠TQR — (1)
In △PQR, ∠PRS is an exterior angle.
∠PRS = ∠QPR + ∠PQR
2∠TRS = ∠QPR + 2∠TQR
2∠TRS – 2∠TQR = ∠QPR
2(∠TRS – ∠TQR) = ∠QPR
Substituting (1) in the above equation:
2∠QTR = ∠QPR
∠QTR = 1/2 ∠QPR
Hence Proved.
📐 Updated Exercise 6.3 Solutions Notice: This page features complete, step-by-step SEBA Class 9 Maths Chapter 6 (Lines and Angles) Exercise 6.3 Solutions updated for the 2026–27 academic session. All geometry proofs, angle sum calculations, and exterior angle property steps strictly follow the latest revised Assam Board (SEBA) textbook.
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