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CBSE Class 9 Mathematics Question 3 of 6

Lines and Angles — Question 5

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Question 5

In Fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

∠ROS = 12\dfrac{1}{2} (∠QOS – ∠POS).

In Fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2 (∠QOS – ∠POS). NCERT Class 9 Mathematics CBSE Solutions.
Answer

Since, OR ⊥ PQ

∴ ∠ROP = 90° and ∠ROQ = 90°

∴ ∠ROS = 90° - ∠POS .......(1)

⇒ ∠QOS = ∠QOR + ∠ROS

⇒ ∠QOS = 90° + ∠ROS

⇒ 90° = ∠QOS - ∠ROS .....(2)

Substituting value of 90° from equation (2) in equation (1), we get :

⇒ ∠ROS = (∠QOS - ∠ROS) - ∠POS

⇒ ∠ROS + ∠ROS = ∠QOS - ∠POS

⇒ 2(∠ROS) = ∠QOS - ∠POS

⇒ ∠ROS = 12\dfrac{1}{2} (∠QOS - ∠POS)

Hence, proved ∠ROS = 12\dfrac{1}{2} (∠QOS - ∠POS).

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Mathematics | Lines and AnglesWeb Content • Interactive Notes

Lines and Angles — Interactive Study Guide

Angle Pair Cheat Sheet

PairRelationshipCondition
ComplementarySum = 90°Any two angles
SupplementarySum = 180°Any two angles
Linear PairSum = 180°Adjacent + on a line
Vertically OppositeEqualIntersecting lines
CorrespondingEqualParallel lines + transversal
Alternate InteriorEqualParallel lines + transversal
Co-interiorSum = 180°Parallel lines + transversal

Triangle Angle Facts

∠A + ∠B + ∠C = 180° (angle sum property)
Exterior angle = sum of remote interior angles

Quick Self-Check

  1. Two parallel lines cut by a transversal: one angle is 72°. Find all 8 angles. (72°, 108° alternating)
  2. In ΔABC, ∠A = 45°, ∠B = 65°. Find ∠C. (70°)
  3. An exterior angle of a triangle is 130°. One non-adjacent interior angle is 50°. Find the other. (80°)

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