MathsCBSE Class 9

Lines and Angles Class 9: Angle Pairs, Parallel Lines and Theorems

Updated 15 October 2026 · Fact-checked

Lines and Angles in Class 9 covers rays, angle measures and angle pairs, the linear pair axiom and its converse, vertically opposite angles, and the corresponding, alternate and co-interior angles made by a transversal with parallel lines. You prove results, including by contradiction, and use them to find unknown angles.

What Lines and Angles is about

The CBSE 2026-27 Class 9 Mathematics syllabus lists Lines and Angles in Unit IV, Geometry. *Ganita Manjari* does not have a chapter on it, so this page follows the CBSE syllabus wording: rays and angles, measures of angles, intersecting lines, pairs of angles, theorems on intersecting lines and on parallel lines, with the linear pair theorem and its converse, proof by contradiction, vertically opposite angles, corresponding, alternate and interior angles, and the transitivity of parallel lines.

The ideas build on Grade 7 and 8 work, but now each result is proved:

when a ray stands on a line, the two angles add to 180° (linear pair), and the converse;
when two lines cross, vertically opposite angles are equal;
when a transversal cuts two parallel lines, corresponding angles are equal, alternate angles are equal and co-interior angles add to 180°; and each converse tests for parallel lines;
lines parallel to the same line are parallel to each other.

These results are the tools for every later geometry chapter (quadrilaterals, circles, triangles). Expect 1-mark angle MCQs, 2- and 3-mark 'find x' questions with reasons, and 5-mark proofs, often through a figure with parallel lines.

Weightage: This chapter is part of Unit IV, Geometry, which carries 25 of the 80 marks in the school annual exam, shared with Introduction to Euclid's Geometry: Axioms and Postulates, Triangles – Congruence Theorems, 4-gons (Quadrilaterals) and Circles.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit IV Geometry

Key concepts

Rays, angles and their measures

A line extends without end in both directions; a line segment has two end points; a ray has one end point and extends without end in one direction. An angle is formed by two rays from a common end point (the vertex); the rays are its arms.

Type	Measure
acute	between 0° and 90°
right	exactly 90°
obtuse	between 90° and 180°
straight	exactly 180°
reflex	between 180° and 360°
complete	360°

Every angle less than 360° has a reflex partner: an angle of $θ$ and its reflex angle add to 360°.

Worked example

Classify 35°, 90°, 145°, 180° and 250°. What is the reflex angle of 145°?

Try it first, then show the solutionSolution

35°: acute; 90°: right; 145°: obtuse; 180°: straight; 250°: reflex.
Reflex angle of 145° $= 360° - 145° = 215°$ .

Answer: Acute, right, obtuse, straight, reflex; the reflex of 145° is 215°.

Pairs of angles

Complementary: two angles adding to 90°.
Supplementary: two angles adding to 180°.
Adjacent: a common vertex and a common arm, with the other arms on opposite sides of the common arm.
Linear pair: adjacent angles whose other arms form a straight line (so they add to 180°).
Vertically opposite: the non-adjacent pairs of angles formed when two lines intersect.

Complementary or supplementary angles need not be adjacent; a linear pair must be adjacent.

Worked example

Find the complement of 37°, the supplement of 128°, and an angle that is equal to its own supplement.

Try it first, then show the solutionSolution

Complement of 37° $= 90° - 37° = 53°$ .
Supplement of 128° $= 180° - 128° = 52°$ .
If $x = 180° - x$ , then $2 x = 180°$ and $x = 90°$ .

Answer: 53°, 52° and 90°.

Linear pair axiom and its converse

Linear Pair Axiom: if a ray stands on a line, the sum of the two adjacent angles so formed is 180°.

Converse: if the sum of two adjacent angles is 180°, their non-common arms form a straight line.

Together they are both directions of one fact, so 'linear pair' and 'adjacent angles adding to 180°' mean the same thing. Use the converse to prove that three points are collinear.

Worked example

Ray OC stands on line AB, with $∠ A O C = (3 x + 20) °$ and $∠ C O B = (2 x + 10) °$ . Find both angles.

Try it first, then show the solutionSolution

Linear pair: $3 x + 20 + 2 x + 10 = 180$ .
$5 x + 30 = 180$ , so $x = 30$ .
$∠ A O C = 110°$ and $∠ C O B = 70°$ .

Answer: 110° and 70°

Vertically opposite angles are equal

Theorem: if two lines intersect, the vertically opposite angles are equal.

*Proof:* lines AB and CD meet at O. Ray OA stands on line CD, so $∠ A O C + ∠ A O D = 180°$ (linear pair). Ray OD stands on line AB, so $∠ A O D + ∠ B O D = 180°$ . Both sums equal 180°, so $∠ A O C + ∠ A O D = ∠ A O D + ∠ B O D$ . Subtracting $∠ A O D$ from both sides gives $∠ A O C = ∠ B O D$ . Similarly $∠ A O D = ∠ B O C$ .

So when two lines cross, the four angles come in two equal pairs, and neighbouring angles are supplementary.

Worked example

Two lines intersect and one of the angles formed is 40°. Find the other three.

Try it first, then show the solutionSolution

The vertically opposite angle is also 40°.
Each neighbouring angle forms a linear pair with 40°: $180° - 40° = 140°$ .
Angles: 40°, 140°, 40°, 140° (they add to 360°).

Answer: 40°, 140°, 140°

Angles at a point

The angles all the way round a point add up to 360° (a complete angle). This follows from the linear pair axiom: extend one ray backwards to make a straight line; the angles on each side of it add to 180°, so all of them add to 360°.

Use it when several rays come out of one point, or for reflex angles.

Worked example

Five rays from a point O make angles $x$ , $2 x$ , $3 x$ , $4 x$ and $5 x$ in turn. Find x and the largest angle.

Try it first, then show the solutionSolution

$x + 2 x + 3 x + 4 x + 5 x = 360°$ .
$15 x = 360°$ , so $x = 24°$ .
Largest $= 5 x = 120°$ .

Answer: $x = 24°$ ; the largest angle is 120°.

Parallel lines and a transversal

A transversal is a line that cuts two (or more) lines at distinct points, making eight angles. When the two lines are parallel:

Corresponding angles are equal (Corresponding Angles Axiom), e.g. both 'top-right' angles;
Alternate interior angles are equal (a 'Z' shape), and alternate exterior angles are equal;
Co-interior angles (interior angles on the same side, a 'C' shape) add to 180°.

The alternate-angle and co-interior results are proved from the corresponding angles axiom using vertically opposite angles and linear pairs.

Worked example

AB ∥ CD and a transversal makes $∠1 = 65°$ at AB (above AB, right of the transversal). Find the corresponding angle at CD, the alternate interior angle to the angle below $∠1$ , and the co-interior partner of that angle.

Try it first, then show the solutionSolution

Corresponding angle at CD $= 65°$ (corresponding angles axiom).
The angle below $∠1$ (right of the transversal, below AB) is $180° - 65° = 115°$ (linear pair).
Its alternate interior angle at CD (left of the transversal, above CD) $= 115°$ .
Its co-interior partner (right of the transversal, above CD) $= 180° - 115° = 65°$ .

Answer: 65°, 115° and 65°.

Tests for parallel lines (converses)

Each property has a true converse, so it can be used as a test. Two lines cut by a transversal are parallel if:

a pair of corresponding angles are equal (Converse of the Corresponding Angles Axiom), or
a pair of alternate interior angles are equal, or
a pair of co-interior angles add to 180°.

Transitivity: lines that are parallel to the same line are parallel to each other. *Reason:* a transversal makes equal corresponding angles with each of them and the third line, so they make equal corresponding angles with each other.

Worked example

A transversal cuts lines l and m. The co-interior angles on one side are $(4 x + 15) °$ and $(3 x + 25) °$ . Find x if l ∥ m.

Try it first, then show the solutionSolution

Parallel lines need co-interior angles to add to 180°.
$7 x + 40 = 180$ , so $x = 20$ .
The angles are 95° and 85°.

Answer: $x = 20$ (angles 95° and 85°).

Proof by contradiction

To prove a statement by contradiction: assume it is false, reason correctly, and reach something impossible (it contradicts a known fact, an axiom or the assumption). Then the assumption must be wrong, so the statement is true.

The CBSE syllabus lists this method among the competencies. Classic uses here: two distinct lines meet in at most one point; two lines perpendicular to the same line cannot meet; a triangle cannot have two right angles.

Worked example

Prove by contradiction that a triangle cannot have two obtuse angles.

Try it first, then show the solutionSolution

Assume a triangle has two obtuse angles, each more than 90°.
Their sum is then more than 180°.
But the three angles of a triangle add to exactly 180°, and the third angle is positive.
This is impossible, so a triangle cannot have two obtuse angles.

Answer: Proved: the assumption leads to an angle sum above 180°.

Angle sum and exterior angle of a triangle

A key theorem on parallel lines: the angles of a triangle add up to 180°.

*Proof:* through vertex A of $△ A B C$ , draw line PQ parallel to BC. Then $∠ P A B = ∠ B$ and $∠ Q A C = ∠ C$ (alternate angles). The three angles at A along line PQ add to 180°: $∠ P A B + ∠ B A C + ∠ Q A C = 180°$ , so $∠ B + ∠ A + ∠ C = 180°$ .

Exterior angle property: an exterior angle of a triangle equals the sum of the two interior opposite angles (it is supplementary to the adjacent interior angle, which is also supplementary to the sum of the other two). So it is greater than each of them.

Worked example

In $△ A B C$ , side BC is extended to D and $∠ A C D = 115°$ . If $∠ A = 50°$ , find $∠ B$ and $∠ A C B$ .

Try it first, then show the solutionSolution

Exterior angle: $∠ A C D = ∠ A + ∠ B$ , so $115° = 50° + ∠ B$ and $∠ B = 65°$ .
$∠ A C B = 180° - 115° = 65°$ (linear pair).
Check: $50° + 65° + 65° = 180°$ .

Answer: $∠ B = 65°$ and $∠ A C B = 65°$ .

Maths in real life: Lines and Angles

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Linear pair

Suppose a ladder leans against a wall and makes an angle of 65° with the ground on one side.

What angle does it make with the ground on the other side?

Try it first, then show the solutionSolution

The ladder is a ray standing on the straight ground.
The two angles form a linear pair: $180° - 65° = 115°$ .

Answer: 115°

Memory hook: A ray on a line splits 180°.

Vertically opposite angles

Suppose the two blades of a pair of scissors make an angle of 30° when opened.

What is the angle between the handles?

Try it first, then show the solutionSolution

The blades and handles are two crossing straight lines.
The handles' angle is vertically opposite the blades' angle.

Answer: 30°

Memory hook: Scissors show opposite angles are twins.

Parallel lines and transversals

Suppose a carpenter makes a gate with two parallel horizontal bars and a diagonal brace at 40° to the top bar.

What angle does the brace make with the bottom bar (on the alternate side)?

Try it first, then show the solutionSolution

The brace is a transversal of the two parallel bars.
Alternate interior angles are equal.

Answer: 40°

Memory hook: Z-shape, equal angles.

Test for parallel lines

Suppose you are lining up two shelves and a straight batten across them makes co-interior angles of 88° and 92°.

Are the shelves parallel?

Try it first, then show the solutionSolution

$88° + 92° = 180°$ .
Co-interior angles adding to 180° prove the lines are parallel.

Answer: Yes, the shelves are parallel.

Memory hook: C-shape adds to 180° for parallels.

Topics in this chapter

Lines and Angles formula sheet

Complementary and supplementary: $x + y = 9 0^{\circ}; x + y = 18 0^{\circ}$
Linear pair: $∠ A O C + ∠ C O B = 18 0^{\circ}$; And its converse.
Vertically opposite angles: $∠ A O C = ∠ B O D, ∠ A O D = ∠ B O C$
Angles at a point: $sum = 36 0^{\circ}$
Reflex angle: $reflex angle = 36 0^{\circ} - θ$
Corresponding angles: $l ∥ m \Leftrightarrow corresponding angles are equal$
Alternate interior angles: $l ∥ m \Leftrightarrow alternate interior angles are equal$
Co-interior angles: $l ∥ m \Leftrightarrow ∠1 + ∠2 = 18 0^{\circ}$
Transitivity: $l ∥ n, m ∥ n \Rightarrow l ∥ m$
Angle sum of a triangle: $∠ A + ∠ B + ∠ C = 18 0^{\circ}$
Exterior angle: $exterior angle = sum of the two interior opposite angles$

How the school exam asks Lines and Angles

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Angle pairs	Find complements, supplements, an angle equal to a fraction of its supplement, or angles in a ratio forming a linear pair.	Write the defining equation (sum 90° or 180°), substitute, and solve.
Intersecting lines	Two or three lines through a point with angles in algebraic form; find x and each angle.	Use vertically opposite angles (equal), linear pairs (180°) and angles at a point (360°), giving each reason.
Parallel lines with a transversal	Figures with AB ∥ CD and one or more transversals; find marked angles.	Spot F (corresponding), Z (alternate) and C (co-interior) shapes. When stuck, draw an extra line through a bend parallel to the given lines.
Prove lines are parallel	Given some angles, show that two lines are parallel, or find x for which they are.	Show one of: equal corresponding angles, equal alternate angles, or co-interior angles adding to 180°.
Proofs	Prove that vertically opposite angles are equal, that bisectors of a linear pair are perpendicular, the angle sum property, or a result by contradiction.	Given, To prove, Construction (if any), then statements with reasons; for contradiction, state the assumption clearly and show the impossibility.
Triangle angles	Use the angle sum or the exterior angle property within parallel-line figures.	Find what you can from parallels first, then use 180° in the triangle or the exterior angle shortcut.

Tricks to remember Lines and Angles

Short memory aids. Say each one aloud once, then use it on a question the same day.

F, Z and C
F-shape: corresponding (equal). Z-shape: alternate (equal). C-shape: co-interior (add to 180°).
Use it when: Spotting angle pairs in parallel-line figures.
Crossed lines, opposite twins
Angles opposite each other at a crossing are twins; neighbours are partners adding to 180°.
Use it when: Intersecting-lines questions.
C for Complementary and Corner (90°), S for Supplementary and Straight (180°)
Complementary makes a corner; supplementary makes a straight line.
Use it when: Never mixing up 90° and 180°.
Bend? Draw a parallel
If a path bends between two parallel lines, draw a third parallel through the bend.
Use it when: Harder 'find the angle at the bend' questions.

Lines and Angles: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Supplement · 1 mark · MCQExam level

An angle is 4 times its supplement. The angle is
(a) 36°
(b) 144°
(c) 72°
(d) 120°

Try it first, then show the solutionSolution

$x = 4 (180° - x)$ gives $5 x = 720°$ .
$x = 144°$ .

Answer: (b) 144°

Example 2: Vertically opposite · 1 mark · MCQExam level

Two lines intersect so that one angle is 3 times its neighbour. The smaller angle is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Try it first, then show the solutionSolution

Neighbouring angles form a linear pair: $x + 3 x = 180°$ .
$x = 45°$ .

Answer: (b) 45°

Example 3: Co-interior angles (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): If two parallel lines are cut by a transversal and one co-interior angle is 72°, the other is 108°.
Reason (R): Interior angles on the same side of a transversal to parallel lines are supplementary.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$180° - 72° = 108°$ , exactly by R.

Answer: (a)

Example 4: Linear pair in a ratio · 2 marks · Very short answerExam level

Two angles forming a linear pair are in the ratio 4 : 5. Find them.

Try it first, then show the solutionSolution

$4 x + 5 x = 180°$ , so $x = 20°$ .
Angles 80° and 100°.

Answer: 80° and 100°

Example 5: Three lines through a point · 3 marks · Short answerHarder exam level

Lines AB, CD and EF pass through O. $∠ A O C = 50°$ and $∠ C O E = 70°$ (adjacent, with AB as a straight line). Find $∠ E O B$ , $∠ B O D$ and $∠ D O F$ .

Try it first, then show the solutionSolution

Along line AB: $∠ A O C + ∠ C O E + ∠ E O B = 180°$ , so $∠ E O B = 60°$ .
$∠ B O D = ∠ A O C = 50°$ (vertically opposite).
$∠ D O F = ∠ C O E = 70°$ (vertically opposite).

Answer: 60°, 50° and 70°

Example 6: Bisectors of a linear pair · 3 marks · Short answerHarder exam level

Prove that the bisectors of the two angles of a linear pair are perpendicular to each other.

Try it first, then show the solutionSolution

Let ray OC stand on line AB, with $∠ A O C = 2 a$ and $∠ C O B = 2 b$ , so $2 a + 2 b = 180°$ (linear pair).
The bisectors OX and OY make $∠ X O C = a$ and $∠ C O Y = b$ .
$∠ X O Y = a + b = \frac{180°}{2} = 90°$ .

Answer: The bisectors meet at 90°.

Example 7: Alternate angles with algebra · 3 marks · Short answerHarder exam level

AB ∥ CD and a transversal cuts them at P and Q. The alternate interior angles $∠ A P Q$ and $∠ P Q D$ are $(2 y + 10) °$ and $(3 y - 20) °$ . Find y and $∠ P QC$ .

Try it first, then show the solutionSolution

$2 y + 10 = 3 y - 20$ , so $y = 30$ .
$∠ P Q D = 70°$ .
$∠ P QC = 180° - 70° = 110°$ (linear pair).

Answer: $y = 30$ ; $∠ P QC = 110°$

Example 8: Angle at a bend · 3 marks · Short answerHarder exam level

AB ∥ CD. A point E lies between the two lines, on the same side of BD as A and C, with $∠ A B E = 40°$ and $∠ E D C = 35°$ . Find $∠ B E D$ .

Try it first, then show the solutionSolution

Through E draw EF parallel to AB (and so to CD, by transitivity).
$∠ B E F = ∠ A B E = 40°$ (alternate angles, AB ∥ EF).
$∠ F E D = ∠ E D C = 35°$ (alternate angles, EF ∥ CD).
$∠ B E D = 40° + 35° = 75°$ .

Answer: 75°

Example 9: Proving vertically opposite angles equal · 5 marks · Long answerHarder exam level

Prove that if two lines intersect, the vertically opposite angles are equal. Then find all four angles if one is $(5 x - 10) °$ and its vertically opposite angle is $(3 x + 30) °$ .

Try it first, then show the solutionSolution

Lines AB and CD meet at O. $∠ A O C + ∠ A O D = 180°$ (linear pair, ray OA on CD).
$∠ A O D + ∠ B O D = 180°$ (linear pair, ray OD on AB).
So $∠ A O C + ∠ A O D = ∠ A O D + ∠ B O D$ ; subtracting $∠ A O D$ : $∠ A O C = ∠ B O D$ . Similarly $∠ A O D = ∠ B O C$ .
$5 x - 10 = 3 x + 30$ gives $x = 20$ , so the pair is 90° each.
The other two angles are $180° - 90° = 90°$ : all four angles are right angles (the lines are perpendicular).

Answer: Proved; all four angles are 90°.

Example 10: Two perpendiculars · 3 marks · Short answerHOTS / competency

Lines l and m are both perpendicular to line n. Prove that l ∥ m. Then prove by contradiction that l and m cannot meet.

Try it first, then show the solutionSolution

n is a transversal to l and m; the co-interior angles are $90° + 90° = 180°$ , so l ∥ m.
By contradiction: suppose l and m meet at P. Then P and the two feet on n form a triangle with two right angles.
Its angle sum would exceed 180° once the third angle is added, which is impossible. So they do not meet.

Answer: Proved both ways.

Example 11: Exterior angle in a parallel figure · 4 marks · Case-basedHOTS / competency

In $△ P QR$ , side QR is extended to S, and RT is drawn parallel to QP. If $∠ P = 55°$ and $∠ Q = 65°$ , find $∠ P R T$ and $∠ T R S$ , and use them to prove the exterior angle property for this triangle.

Try it first, then show the solutionSolution

$∠ P R T = ∠ P = 55°$ (alternate angles, RT ∥ QP, transversal PR).
$∠ T R S = ∠ Q = 65°$ (corresponding angles, RT ∥ QP, transversal QS).
Exterior $∠ P R S = ∠ P R T + ∠ T R S = 55° + 65° = 120°$ .
So the exterior angle equals $∠ P + ∠ Q$ , the sum of the interior opposite angles.

Answer: 55° and 65°; the exterior angle is 120° $= ∠ P + ∠ Q$ .

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

A railway crossing

Two straight, parallel railway tracks are crossed by a straight road. On a map, the road makes an angle of 62° with the first track, measured on the north side of the track and the east side of the road.

(i) · 1 mark
What angle does the road make with the second track, in the matching position?
Answer
Answer: 62°
Corresponding angles are equal because the tracks are parallel.
(ii) · 1 mark
Find the angle between the road and the first track on the south side, east of the road.
Answer
Answer: 118°
It forms a linear pair with 62°: $180° - 62° = 118°$ .
(iii) · 2 marks
A straight footpath crosses both tracks. A surveyor reports that the interior angles it makes with the two tracks, on the same side of the footpath, are 110° and 72°. Can both measurements be correct? Explain.
Answer
Answer: No
A transversal of parallel lines makes co-interior angles adding to 180° (1 mark). $110° + 72° = 182° \neq = 180°$ , so the measurements cannot both be correct, or the footpath is not straight (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Two adjacent angles forming a linear pair are in the ratio 4 : 5. Find them. (2 marks)

Step	Marks
$4 x + 5 x = 180°$ (linear pair)	1
$x = 20°$ : the angles are 80° and 100°	1

AB ∥ CD, and a transversal cuts AB at P and CD at Q. $∠ A P Q = (2 y + 10) °$ and $∠ P Q D = (3 y - 20) °$ are alternate interior angles. Find y and $∠ P QC$ . (3 marks)

Step	Marks
Alternate interior angles are equal: $2 y + 10 = 3 y - 20$	1
$y = 30$ , so $∠ P Q D = 70°$	1
$∠ P QC = 180° - 70° = 110°$ (linear pair)	1

Prove that the sum of the angles of a triangle is 180°. (5 marks)

Step	Marks
Given $△ A B C$ . To prove $∠ A + ∠ B + ∠ C = 180°$	1
Construction: through A draw line PQ parallel to BC	1
$∠ P A B = ∠ A B C$ and $∠ Q A C = ∠ A C B$ (alternate angles, PQ ∥ BC)	1
$∠ P A B + ∠ B A C + ∠ Q A C = 180°$ (angles on the straight line PQ)	1
Substituting: $∠ B + ∠ A + ∠ C = 180°$	1

Where students lose marks in Lines and Angles

The slips examiners see most often, and what to write instead.

Mistake: Using parallel-line properties when lines are not given as parallel
Lines that look parallel are assumed to be.
Wrong working
The figure shows two lines that look parallel, so alternate angles are equal.
Right working
Use alternate angles only if the lines are given (or proved) parallel.
Do this: Check the given information, not the picture.
Usually costs: The whole answer.
Mistake: Taking co-interior angles as equal
All transversal pairs are treated alike.
Wrong working
AB ∥ CD, co-interior angles x and 70°, so $x = 70°$ .
Right working
Co-interior angles add to 180°: $x = 110°$ .
Do this: Equal for F and Z shapes; supplementary for C shapes.
Usually costs: 1 to 2 marks.
Mistake: Writing answers without reasons
The angle seems obvious from the figure.
Wrong working
$∠3 = 65°$
Right working
$∠3 = 65°$ (alternate interior angles, AB ∥ CD)
Do this: Write the reason in brackets after every angle you find.
Usually costs: Half the marks in many questions.
Mistake: Confusing complementary and supplementary
The two words sound alike.
Wrong working
Supplement of 40° $= 50°$
Right working
Supplement $= 140°$ ; complement $= 50°$
Do this: Supplementary = straight line (180°).
Usually costs: 1 mark.

Practice questions on Lines and Angles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The complement of 48° is
(a) 132°
(b) 42°
(c) 52°
(d) 312°
Hint
Subtract from 90°.
Answer
(b) 42°
Q2 · 1 mark · MCQExam level
If two parallel lines are cut by a transversal, alternate interior angles are
(a) supplementary
(b) complementary
(c) equal
(d) always 90°
Hint
Z-shape.
Answer
(c) equal
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): If two adjacent angles are 115° and 65°, their non-common arms form a straight line.
Reason (R): If two adjacent angles add to 180°, their non-common arms form a line.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Converse of the linear pair axiom.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Find the angle which is equal to one-fifth of its supplement.
Hint
$x = \frac{1}{5} (180 - x)$ .
Answer
$5 x = 180 - x$ , so $x = 30°$ .
Q5 · 3 marks · Short answerHarder exam level
Four rays from O make angles $(2 x) °$ , $(3 x + 10) °$ , $(4 x - 20) °$ and $(x + 70) °$ in turn. Find x and the angles.
Hint
Angles at a point.
Answer
$10 x + 60 = 360$ , $x = 30$ : 60°, 100°, 100°, 100°.
Q6 · 3 marks · Short answerHarder exam level
AB ∥ CD, and a transversal makes $∠1 = (3 x - 15) °$ and its corresponding angle $(2 x + 25) °$ . Find x and both co-interior angles at that crossing.
Hint
Corresponding angles are equal.
Answer
$3 x - 15 = 2 x + 25$ , $x = 40$ : the angle is 105°; its co-interior partner is $180° - 105° = 75°$ .
Q7 · 2 marks · Very short answerHarder exam level
Line l is parallel to line m, and line m is parallel to line n. A transversal makes an angle of 48° with l. What angle does it make with n (in the corresponding position)? Which property do you use?
Hint
Transitivity.
Answer
48°; l ∥ n by transitivity, so corresponding angles are equal.
Q8 · 3 marks · Short answerHarder exam level
AB ∥ CD. A point E lies between the two lines, on the same side of BD as A and C, with $∠ A B E = 55°$ and $∠ C D E = 45°$ . Find $∠ B E D$ .
Hint
Draw a parallel through E.
Answer
$55° + 45° = 100°$ .
Q9 · 3 marks · Short answerHarder exam level
In $△ A B C$ , $∠ A = 2∠ B$ and $∠ C = 3∠ B$ . Find the angles. Is the triangle right-angled?
Hint
Angle sum.
Answer
$6∠ B = 180°$ : $∠ B = 30°$ , $∠ A = 60°$ , $∠ C = 90°$ . Yes, right-angled at C.
Q10 · 5 marks · Long answerHarder exam level
Prove that if a transversal cuts two lines so that a pair of alternate interior angles are equal, the lines are parallel. (Use the converse of the corresponding angles axiom.)
Hint
Turn the alternate angle into a corresponding angle using vertically opposite angles.
Answer
Let the transversal cut l at P and m at Q with alternate interior angles $∠1$ (at P) and $∠2$ (at Q) equal. The angle vertically opposite $∠1$ at P equals $∠1$ , and it is in the corresponding position to $∠2$ . So a pair of corresponding angles are equal, and by the converse of the corresponding angles axiom, l ∥ m.
Q11 · 3 marks · Short answerHOTS / competency
Prove by contradiction that through a point P not on line l there is at most one line perpendicular to l.
Hint
Two perpendiculars would make a triangle.
Answer
Suppose two different perpendiculars from P meet l at A and B. Then triangle PAB has two right angles (at A and B), so its angle sum exceeds 180°, which is impossible. So there is at most one perpendicular.
Q12 · 4 marks · Case-basedHOTS / competency
A ray of light strikes a flat mirror and reflects so that the angle with the mirror before and after is equal. Two parallel mirrors are placed facing each other. Show that the ray leaving the second mirror is parallel to the ray entering the first.
Hint
Use alternate angles between the parallel mirrors.
Answer
Let the incoming ray make angle a with mirror 1; it leaves at angle a. The ray between the mirrors meets mirror 2 at angle a too (alternate angles, mirrors parallel) and leaves at angle a. The incoming and outgoing rays make equal alternate angles with the middle ray, so they are parallel.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a reflex angle?Show answer
An angle between 180° and 360°.
2.State the linear pair axiom.Show answer
If a ray stands on a line, the two adjacent angles formed add up to 180°.
3.State the converse of the linear pair axiom.Show answer
If two adjacent angles add to 180°, their non-common arms form a line.
4.What can you say about vertically opposite angles?Show answer
They are equal.
5.What do the angles round a point add up to?Show answer
360°.
6.With parallel lines, which angle pairs are equal and which are supplementary?Show answer
Corresponding and alternate angles are equal; co-interior angles are supplementary.
7.What does transitivity of parallel lines say?Show answer
Lines parallel to the same line are parallel to each other.
8.What is proof by contradiction?Show answer
Assuming the statement is false and showing this leads to something impossible.
9.State the exterior angle property.Show answer
An exterior angle of a triangle equals the sum of the two interior opposite angles.

Best order to study it

Revise rays, angles and the types of angles, including reflex angles.
Learn the angle pairs and practise complements and supplements.
Learn the linear pair axiom and its converse, then prove vertically opposite angles are equal.
Learn corresponding, alternate and co-interior angles with parallel lines, and their converses.
Practise multi-step figures, drawing extra parallel lines where needed.
Finish with proofs: angle sum, exterior angle, and simple proofs by contradiction.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find the reflex angle of 75°?
Can you find an angle that is half its supplement?
Can you prove that vertically opposite angles are equal?
Can you find x when three lines meet at a point with angles in terms of x?
Can you name and mark corresponding, alternate and co-interior angles?
Can you show that two lines are parallel from given angles?
Can you find the angle at a bend between two parallel lines?
Can you prove the angle sum property of a triangle?
Can you write a short proof by contradiction?

Lines and Angles: frequently asked questions

Is Lines and Angles in the 2026-27 CBSE Class 9 syllabus without an NCERT chapter?

Yes. CBSE's syllabus lists Lines and Angles in Unit IV, Geometry, but Ganita Manjari Part I and Part II have no chapter on it. This page follows the CBSE syllabus wording; check with your school how it will be taught and examined.

Which results are axioms and which are theorems?

The linear pair axiom and the corresponding angles axiom are taken as starting points (with their converses). Vertically opposite angles, the alternate angle and co-interior angle results, transitivity of parallels and the angle sum property are proved from them.

Do I always have to write reasons?

Yes. In geometry, the reason (for example, 'alternate angles, AB ∥ CD') is usually worth as much as the number.

What is the trick for angles at a bend between two parallel lines?

Draw a third line through the bend, parallel to the given lines. The angle at the bend splits into two parts, each found as an alternate angle; add them.

Based on: CBSE Mathematics syllabus, Class IX 2026-27