CBSE • Class 9 • Mathematics • Chapter 6 (Lines and Angles) • Exercise 6.1

Exercise 6.1: Lines and Angles — NCERT Solutions

Intersecting lines — linear pairs and vertically opposite angles.

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What this exercise covers

Linear pairVertically oppositeAngle addition

Step-by-step solutions — Exercise 6.1

4 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 6.1 Q1 • 3 marks

Two lines ABAB and CDCD intersect at a point OO. If ∠AOC=50∘\angle AOC = 50^\circ, find the measures of ∠BOD\angle BOD, ∠AOD\angle AOD and ∠BOC\angle BOC.
Hint (Socratic — try this first)▾
Which angles are vertically opposite to each other, and which pairs form a linear pair?
Step-by-step solution▾

When two lines intersect, vertically opposite angles are equal.

  • ∠BOD\angle BOD is vertically opposite to ∠AOC\angle AOC, so ∠BOD=50∘\angle BOD = 50^\circ.

Now ∠AOC\angle AOC and ∠AOD\angle AOD form a linear pair, so they add to 180∘180^\circ: ∠AOD=180∘−50∘=130∘\angle AOD = 180^\circ - 50^\circ = 130^\circ

  • ∠BOC\angle BOC is vertically opposite to ∠AOD\angle AOD, so ∠BOC=130∘\angle BOC = 130^\circ.

Answer: ∠BOD=50∘\angle BOD = 50^\circ, ∠AOD=130∘\angle AOD = 130^\circ, ∠BOC=130∘\angle BOC = 130^\circ.

Common mistake:
Assuming all four angles at the point are equal instead of recognising that adjacent angles are supplementary (linear pair).
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Exercise 6.1 Q2 • 3 marks

In the figure, ray OPOP stands on line LMLM. If ∠LOP=(2x+15)∘\angle LOP = (2x + 15)^\circ and ∠MOP=(3x−25)∘\angle MOP = (3x - 25)^\circ, find the value of xx and both angles.
Hint (Socratic — try this first)▾
What is the sum of the two angles a ray makes on a straight line?
Step-by-step solution▾

Since ray OPOP stands on the straight line LMLM, ∠LOP\angle LOP and ∠MOP\angle MOP form a linear pair: ∠LOP+∠MOP=180∘\angle LOP + \angle MOP = 180^\circ (2x+15)+(3x−25)=180(2x+15) + (3x-25) = 180 5x−10=1805x - 10 = 180 5x=190  ⟹  x=385x = 190 \implies x = 38

Now substitute:

  • ∠LOP=2(38)+15=91∘\angle LOP = 2(38) + 15 = 91^\circ
  • ∠MOP=3(38)−25=89∘\angle MOP = 3(38) - 25 = 89^\circ

Check: 91∘+89∘=180∘91^\circ + 89^\circ = 180^\circ ✓

Common mistake:
Setting the two expressions equal to each other instead of adding them to 180∘180^\circ.
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Exercise 6.1 Q3 • 2 marks

The two adjacent angles on a straight line are in the ratio 4:54 : 5. Find both angles.
Hint (Socratic — try this first)▾
If the angles are 4k4k and 5k5k, what must their sum be?
Step-by-step solution▾

Let the two adjacent angles be 4k4k and 5k5k. Since they lie on a straight line (linear pair): 4k+5k=180∘4k + 5k = 180^\circ 9k=180∘  ⟹  k=20∘9k = 180^\circ \implies k = 20^\circ

Therefore:

  • First angle =4k=80∘= 4k = 80^\circ
  • Second angle =5k=100∘= 5k = 100^\circ

Answer: 80∘80^\circ and 100∘100^\circ.

Common mistake:
Dividing 180∘180^\circ by 2 and applying the ratio incorrectly, instead of dividing by the total number of ratio parts (9).
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Exercise 6.1 Q4 • 4 marks

Three lines pass through a common point OO. If ∠1=40∘\angle 1 = 40^\circ and ∠2=65∘\angle 2 = 65^\circ are two adjacent angles around OO, and the remaining angles around the point are ∠3,∠4,∠5,∠6\angle 3, \angle 4, \angle 5, \angle 6, prove that the sum of all angles formed around point OO is 360∘360^\circ and are the vertically opposite pairs consistent with the given values.
Hint (Socratic — try this first)▾
How many degrees make one complete turn around a point, and which angles are equal by the vertically opposite angle property?
Step-by-step solution▾

The angles around a point add up to a complete rotation: ∠1+∠2+∠3+∠4+∠5+∠6=360∘\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 = 360^\circ

Since three lines cross at OO, angles occur in vertically opposite pairs:

  • ∠1\angle 1 and its opposite are equal =40∘= 40^\circ
  • ∠2\angle 2 and its opposite are equal =65∘= 65^\circ

The third pair of opposite angles: on one side of a line, ∠1+∠2+third angle=180∘\angle 1 + \angle 2 + \text{third angle} = 180^\circ (angles on a straight line). third angle=180∘−40∘−65∘=75∘\text{third angle} = 180^\circ - 40^\circ - 65^\circ = 75^\circ

So the six angles are 40∘,65∘,75∘,40∘,65∘,75∘40^\circ, 65^\circ, 75^\circ, 40^\circ, 65^\circ, 75^\circ.

Sum =2(40∘+65∘+75∘)=2×180∘=360∘= 2(40^\circ + 65^\circ + 75^\circ) = 2 \times 180^\circ = 360^\circ ✓

Hence the vertically opposite pairs are consistent and the total is 360∘360^\circ.

Common mistake:
Forgetting that angles around a point sum to 360∘360^\circ (not 180∘180^\circ) and mismatching vertically opposite pairs.
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How to approach Exercise 6.1

  1. Re-read the chapter summary first. Open Lines and Angles and refresh the key concepts: Linear pair, Vertically opposite angles, Parallel lines, Transversal.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Lines and Angles

  1. Exercise 6.1 — Intersecting lines — linear pairs and vertically opposite angles.
  2. Exercise 6.2 — Parallel lines and a transversal — corresponding, alternate and co-interior angles.
  3. Exercise 6.3 — Angle-sum property of a triangle and exterior angle theorem applications.

Exercise 6.2 →

Parallel lines and a transversal — corresponding, alternate and co-interior angles.

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