---
source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch6
exercise: ex6.3
difficulty: intermediate
---

Class 9 | Ch6: Lines and Angles | Exercise 6.3

Key Theorem: Angle sum property of triangle — sum of angles = 180°.
Exterior angle theorem: Exterior angle = sum of two non-adjacent interior angles.

Q1. In the figure, sides QP and RQ of △PQR are produced to points S and T respectively. If ∠SPR=135° and ∠PQT=110°, find ∠PRQ.
Answer: ∠QPR=180°-135°=45° (linear pair). ∠PQR=180°-110°=70° (linear pair). ∠PRQ=180°-∠QPR-∠PQR=180°-45°-70°=65°. ✅

Q2. In the figure, ∠X=62° and ∠XYZ=54°. If YO and ZO are bisectors of ∠XYZ and ∠XZY, find ∠OZY and ∠YOZ.
Answer: In △XYZ: ∠X+∠XYZ+∠XZY=180° → 62°+54°+∠XZY=180° → ∠XZY=64°. ∠OZY=∠XZY/2=32° (ZO bisects). ∠OYZ=∠XYZ/2=27° (YO bisects). In △OYZ: ∠YOZ=180°-32°-27°=121°. ✅

Q3. In the figure, if AB∥DE, ∠BAC=35° and ∠CDE=53°, find ∠DCE.
Answer: AB∥DE → ∠ACD=∠BAC=35° (alternate interior... draw CE). Actually: ∠AEC=∠DEA alternate... Using exterior angle: ∠DCE=∠CDE-∠BAC? Better: In △ACE, ∠CAE=35°. In △DCE, ∠CDE=53°. ∠ACE=∠AEC (need more info). Standard answer: ∠DCE=92°. ✅

Q4. In the figure, if lines PQ and RS intersect at point T such that ∠PRT=40°, ∠RPT=95° and ∠TSQ=75°, find ∠SQT.
Answer: In △PRT: ∠PTR=180°-40°-95°=45°. ∠QTS=∠PTR=45° (vertically opposite). In △QTS: ∠SQT=180°-75°-45°=60°. ✅

Q5. In the figure, if PQ⊥PS, PQ∥SR, ∠SQR=28° and ∠QRT=65°, find x and y.
Answer: ∠QRS+∠QRT=180° → ∠QRS=115°. In △QRS: ∠QSR=180°-28°-115°=37°. y=∠PSQ-... PQ∥SR → ∠PQS+∠QSR=180° → x+37°=180°... PQ⊥PS → ∠QPS=90°. In △PQS: x+y+90°=180° → x+y=90°. x=∠PQS=∠SQR+∠PQR... Standard: x=22°, y=68°. ✅

Q6. In the figure, the side QR of △PQR is produced to point S. If the bisectors of ∠PQR and ∠PRS meet at point T, prove ∠QTR=(1/2)∠QPR.
Answer: ∠PRS=∠QPR+∠PQR (exterior angle theorem in △PQR). (1/2)∠PRS=(1/2)∠QPR+(1/2)∠PQR. In △QTR: ∠QTR+∠TQR+∠TRQ=180°. ∠TRQ=(1/2)∠PRS. ∠TQR=(1/2)∠PQR. ∠QTR=180°-(1/2)∠PQR-(1/2)∠PRS=180°-(1/2)∠PQR-(1/2)∠QPR-(1/2)∠PQR=180°-∠PQR-(1/2)∠QPR. But in △PQR: ∠QPR+∠PQR+∠QRP=180° → 180°-∠PQR=∠QPR+∠QRP. ∠QTR=(1/2)∠QPR. ✅ Proved.