MathsCBSE Class 9 · Lines and Angles

Linear Pairs, Vertically Opposite Angles and Angles at a Point

Updated 17 October 2026 · Fact-checked

If a ray stands on a line, the two adjacent angles it makes add up to $180^\circ$ (linear pair axiom). When two lines intersect, vertically opposite angles are equal, and the four angles round the point add up to $360^\circ$. Complementary angles add to $90^\circ$ and supplementary angles add to $180^\circ$. Most sums set up one equation from these facts, and proofs quote the fact used at each line.

Same-coloured opposite angles are equal, and any two neighbouring angles (60° + 120°) form a linear pair adding to 180°.

Illustration drawn for QuestionBuddy.

Understand Linear Pairs, Vertically Opposite Angles and Angles at a Point

Types of angles and pairs. An angle is formed by two rays with a common end point (the vertex). Two angles are complementary if they add up to $9 0^{\circ}$ and supplementary if they add up to $18 0^{\circ}$ . Two angles are adjacent if they share a vertex and one arm, and lie on opposite sides of that common arm.

Linear pair axiom. If a ray stands on a line, the sum of the two adjacent angles so formed is $18 0^{\circ}$ . The converse also holds: if two adjacent angles add up to $18 0^{\circ}$ , their non-common arms form a straight line. A linear pair is therefore a pair of adjacent angles whose non-common arms are opposite rays. The angles $∠ A O C$ and $∠ B O C$ on a line $A B$ with $O C$ standing on it are a linear pair, so $∠ A O C + ∠ B O C = 18 0^{\circ}$ .

Vertically opposite angles. When two lines $A B$ and $C D$ intersect at $O$ , the angles $∠ A O C$ and $∠ B O D$ are vertically opposite, and so are $∠ A O D$ and $∠ B O C$ . They are equal. Proof: $∠ A O C + ∠ A O D = 18 0^{\circ}$ and $∠ A O D + ∠ B O D = 18 0^{\circ}$ (linear pairs), so $∠ A O C = ∠ B O D$ . The proof shows why: both are supplements of the same angle.

Angles at a point. All angles formed round a point add up to $36 0^{\circ}$ . If a point has four angles, as in the cross of two lines, they add to $36 0^{\circ}$ and are in two equal pairs. A reflex angle is an angle between $18 0^{\circ}$ and $36 0^{\circ}$ .

Bisectors. If $O X$ bisects $∠ A O C$ , then $∠ A O X = ∠ X O C = \frac{1}{2} ∠ A O C$ . Two useful results: the bisectors of the two angles of a linear pair are perpendicular, because $\frac{1}{2} (∠ A O C + ∠ B O C) = 9 0^{\circ}$ ; and if a ray bisects an angle, its opposite ray bisects the vertically opposite angle.

Solving a sum. (1) Identify what kind of pair each given angle forms: linear pair, vertically opposite, or angles at a point. (2) Write the equation: "sum $= 18 0^{\circ}$ ", "equal", or "sum $= 36 0^{\circ}$ ". (3) Solve and substitute back to get every angle. (4) Check that each linear pair adds to $18 0^{\circ}$ . For ratio problems, write the angles as $ak$ and $bk$ and use the total: a linear pair in the ratio $7 : 5$ has parts totalling $12$ , so the angles are $\frac{180}{12} \times 7 = 10 5^{\circ}$ and $7 5^{\circ}$ .

Clock angles. The hour hand moves $3 0^{\circ}$ per hour, so $0. 5^{\circ}$ per minute; the minute hand moves $6^{\circ}$ per minute. At $3 : 40$ the hour hand is at $3 \times 30 + 40 \times 0.5 = 11 0^{\circ}$ and the minute hand at $24 0^{\circ}$ , so the angle between them is $13 0^{\circ}$ .

Linear Pairs, Vertically Opposite Angles and Angles at a Point: formulas to remember

Complementary and supplementary: $α + β = 9 0^{\circ} or 18 0^{\circ}$; Complementary and supplementary respectively.
Linear pair: $∠ A O C + ∠ B O C = 18 0^{\circ}$; $O C$ stands on line $A B$ .
Vertically opposite angles: $∠ A O C = ∠ B O D, ∠ A O D = ∠ B O C$; Lines $A B$ and $C D$ meet at $O$ .
Angles at a point: $θ_{1} + θ_{2} + \dots = 36 0^{\circ}$; All angles round a point.
Bisectors of a linear pair: $∠ X O Y = 9 0^{\circ}$; $O X$ , $O Y$ bisect the two angles of a linear pair.
Clock hands: $hour hand 0. 5^{\circ} / min, minute hand 6^{\circ} / min$; Take the smaller of $d$ and $36 0^{\circ} - d$ .

How to solve Linear Pairs, Vertically Opposite Angles and Angles at a Point questions

Turn each fact in the figure into one equation, then solve.

1Name each pair of angles: linear pair, vertically opposite, or round a point.
2Write the matching equation: sum $= 18 0^{\circ}$ , equal, or sum $= 36 0^{\circ}$ .
3For ratios, use $k$ : angles $ak$ and $bk$ with total known.
4Solve for the unknown and substitute to find every angle asked.
5Check each linear pair adds to $18 0^{\circ}$ and the angles at the point add to $36 0^{\circ}$ .
6In proofs, write the reason (linear pair, vertically opposite, bisector) beside each line.

Linear Pairs, Vertically Opposite Angles and Angles at a Point in real life

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.

Angles at a point

Suppose a pizza is cut from its centre into 8 equal slices, and a guest takes 3 slices that are next to each other.

What angle do the taken slices make at the centre, and what angle do the remaining slices make?

Try it first, then show the solutionSolution

Each slice makes $\frac{36 0 ^{\circ}}{8} = 4 5^{\circ}$ at the centre.
Three slices: $3 \times 4 5^{\circ} = 13 5^{\circ}$ .
The remaining five slices make $5 \times 4 5^{\circ} = 22 5^{\circ}$ , and $13 5^{\circ} + 22 5^{\circ} = 36 0^{\circ}$ .

Answer: The taken slices make $13 5^{\circ}$ and the rest make $22 5^{\circ}$ (a reflex angle).

Memory hook: All angles round a point make a full turn.

Tricks to remember Linear Pairs, Vertically Opposite Angles and Angles at a Point

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

Straight line: 180
Anything that sits on a straight line adds to 180 degrees; that is the linear pair.
Use it when: Setting up equations.
Across the X: equal
Angles opposite each other across the crossing point are equal.
Use it when: Vertically opposite angles.
Full turn: 360
Angles round a point make a full turn.
Use it when: Several rays from one point.

Linear Pairs, Vertically Opposite Angles and Angles at a Point: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Vertically opposite angles with algebra · 3 marks · Short answerBoard level

Two straight lines $A B$ and $C D$ intersect at $O$ . $∠ A O C = (5 x - 12)^{\circ}$ and $∠ B O D = (3 x + 16)^{\circ}$ . Find $x$ and all four angles.

Try it first, then show the solutionSolution

$∠ A O C$ and $∠ B O D$ are vertically opposite, so $5 x - 12 = 3 x + 16$ . (1)
$2 x = 28$ , so $x = 14$ and $∠ A O C = ∠ B O D = 5 8^{\circ}$ . (1)
$∠ A O D = ∠ B O C = 18 0^{\circ} - 5 8^{\circ} = 12 2^{\circ}$ (linear pair). (1)

Answer: $x = 14$ ; the angles are $5 8^{\circ}, 12 2^{\circ}, 5 8^{\circ}, 12 2^{\circ}$ .

Example 2: Linear pair in a ratio · 3 marks · Short answerBoard level

The angles of a linear pair are in the ratio $7 : 5$ . Find them, and the angle between the bisector of the larger angle and the common arm.

Try it first, then show the solutionSolution

Let the angles be $7 k$ and $5 k$ ; $7 k + 5 k = 18 0^{\circ}$ , so $k = 1 5^{\circ}$ . (1)
The angles are $10 5^{\circ}$ and $7 5^{\circ}$ . (1)
The bisector of the $10 5^{\circ}$ angle makes $52. 5^{\circ}$ with the common arm. (1)

Answer: The angles are $10 5^{\circ}$ and $7 5^{\circ}$ ; the bisector makes $52. 5^{\circ}$ with the common arm.

Example 3: Opposite ray bisects the opposite angle · 5 marks · Long answerHarder board level

Lines $A B$ and $C D$ intersect at $O$ . Ray $O E$ bisects $∠ A O C$ and $O F$ is the ray opposite to $O E$ . Prove that $O F$ bisects $∠ B O D$ , and find $∠ B O F$ if $∠ A O C = 6 4^{\circ}$ .

Try it first, then show the solutionSolution

Let $∠ A O E = ∠ E O C = x$ since $O E$ bisects $∠ A O C$ . (1)
$A B$ and $E F$ are straight lines meeting at $O$ , so $∠ B O F = ∠ A O E = x$ (vertically opposite angles). (1)
$C D$ and $E F$ are straight lines meeting at $O$ , so $∠ D O F = ∠ C O E = x$ (vertically opposite angles). (1)
Therefore $∠ B O F = ∠ D O F = x$ , so $O F$ bisects $∠ B O D$ . (1)
With $∠ A O C = 6 4^{\circ}$ , $x = 3 2^{\circ}$ , so $∠ B O F = 3 2^{\circ}$ . (1)

Answer: $O F$ bisects $∠ B O D$ , and $∠ B O F = 3 2^{\circ}$ .

Example 4: Four angles at a point · 3 marks · Short answerHarder board level

Four angles at a point $O$ are $(x + 20)^{\circ}$ , $(2 x - 10)^{\circ}$ , $(3 x + 30)^{\circ}$ and $(4 x)^{\circ}$ . Find $x$ and the angles, and say which two of them form a linear pair if the first and the last are adjacent.

Try it first, then show the solutionSolution

Angles at a point add to $36 0^{\circ}$ : $10 x + 40 = 360$ . (1)
$x = 32$ , so the angles are $5 2^{\circ}, 5 4^{\circ}, 12 6^{\circ}, 12 8^{\circ}$ (sum $36 0^{\circ}$ ). (1)
$5 2^{\circ} + 12 8^{\circ} = 18 0^{\circ}$ , so if these two angles are adjacent they form a linear pair (and so do $5 4^{\circ}$ and $12 6^{\circ}$ ). (1)

Answer: $x = 32$ ; angles $5 2^{\circ}, 5 4^{\circ}, 12 6^{\circ}, 12 8^{\circ}$ ; the pairs adding to $18 0^{\circ}$ are $(5 2^{\circ}, 12 8^{\circ})$ and $(5 4^{\circ}, 12 6^{\circ})$ .

Example 5: Road junction · 4 marks · Case-basedHOTS / competency

Case: two straight roads $A B$ and $C D$ cross at a junction $O$ , with $∠ A O D = 11 8^{\circ}$ . A footpath $O E$ is built inside $∠ A O C$ so that it bisects it. (i) Find $∠ A O C$ . (ii) Find $∠ B O C$ . (iii) Find $∠ A O E$ . (iv) Find $∠ E O D$ and $∠ E O B$ .

Try it first, then show the solutionSolution

(i) $∠ A O C$ and $∠ A O D$ are a linear pair: $∠ A O C = 18 0^{\circ} - 11 8^{\circ} = 6 2^{\circ}$ . (1)
(ii) $∠ B O C$ is vertically opposite to $∠ A O D$ , so $∠ B O C = 11 8^{\circ}$ . (1)
(iii) $∠ A O E = \frac{6 2 ^{\circ}}{2} = 3 1^{\circ}$ . (1)
(iv) $∠ E O C = 3 1^{\circ}$ and $C D$ is a line, so $∠ E O D = 18 0^{\circ} - 3 1^{\circ} = 14 9^{\circ}$ ; likewise $∠ E O B = 18 0^{\circ} - 3 1^{\circ} = 14 9^{\circ}$ . (1)

Answer: (i) $6 2^{\circ}$ ; (ii) $11 8^{\circ}$ ; (iii) $3 1^{\circ}$ ; (iv) both $14 9^{\circ}$ .

Where students lose marks in Linear Pairs, Vertically Opposite Angles and Angles at a Point

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

Mistake: Using 90 instead of 180 for a linear pair
Confusing complementary with supplementary.
Wrong working
$x + 2 x = 90$ for a linear pair.
Right working
$x + 2 x = 180$ .
Do this: Straight line means 180 degrees.
Usually costs: The whole answer.
Mistake: Not giving the reason
Equations without the property named.
Wrong working
$5 x - 12 = 3 x + 16$ .
Right working
$5 x - 12 = 3 x + 16$ (vertically opposite angles are equal).
Do this: Write the property in brackets.
Usually costs: ½ to 1 mark.
Mistake: Wrong pair of vertically opposite angles
Calling adjacent angles vertically opposite.
Wrong working
$∠ A O C = ∠ A O D$ .
Right working
$∠ A O C = ∠ B O D$ .
Do this: Vertically opposite angles share only the vertex.
Usually costs: 1 mark.

Exam tips

Mark the figure with the given values and the equal angles before writing equations.
Write the property name next to each equation.
After solving, substitute back to check each linear pair.
For clock questions, always take the smaller angle (at most $18 0^{\circ}$ ).

Practice questions on Linear Pairs, Vertically Opposite Angles and Angles at a Point

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 · MCQBoard level
Two angles are complementary and one is four times the other. The larger angle is
(a) $1 8^{\circ}$
(b) $5 4^{\circ}$
(c) $7 2^{\circ}$
(d) $3 6^{\circ}$
Hint
Let the angles be $x$ and $4 x$ ; they add to $9 0^{\circ}$ .
Answer
(c) $7 2^{\circ}$ . $5 x = 9 0^{\circ}$ gives $x = 1 8^{\circ}$ , so the larger angle is $4 \times 1 8^{\circ} = 7 2^{\circ}$ .
Q2 · 3 marks · Short answerBoard level
Two lines $A B$ and $C D$ intersect at $O$ . If $∠ A O C + ∠ B O D = 11 2^{\circ}$ , find all four angles at $O$ , naming the property used.
Hint
Vertically opposite angles are equal.
Answer
$∠ A O C = ∠ B O D$ (vertically opposite), so each is $5 6^{\circ}$ . $∠ A O D = 18 0^{\circ} - 5 6^{\circ} = 12 4^{\circ}$ (linear pair) and $∠ B O C = 12 4^{\circ}$ (vertically opposite to it).
Q3 · 3 marks · Short answerHarder board level
The difference between two supplementary angles is $2 6^{\circ}$ . Find the angles, and find the complement of the smaller one.
Hint
Let the angles be $x$ and $18 0^{\circ} - x$ .
Answer
Larger $= \frac{180 + 26}{2} = 10 3^{\circ}$ , smaller $= 7 7^{\circ}$ (check: $103 + 77 = 180$ , $103 - 77 = 26$ ). Complement of $7 7^{\circ}$ is $9 0^{\circ} - 7 7^{\circ} = 1 3^{\circ}$ .
Q4 · 5 marks · Long answerHarder board level
Prove that the bisectors of the two angles forming a linear pair are at right angles to each other. If one of the angles is $5 0^{\circ}$ , find the angles the bisectors make with the line.
Hint
Write half of the sum of the linear pair.
Answer
Let $O C$ stand on line $A B$ , $∠ A O C + ∠ B O C = 18 0^{\circ}$ . $O X$ bisects $∠ A O C$ and $O Y$ bisects $∠ B O C$ . Then $∠ X O Y = ∠ X O C + ∠ C O Y = \frac{1}{2} ∠ A O C + \frac{1}{2} ∠ B O C = \frac{1}{2} (18 0^{\circ}) = 9 0^{\circ}$ . With $∠ A O C = 5 0^{\circ}$ : $∠ A O X = ∠ X O C = 2 5^{\circ}$ , $∠ B O C = 13 0^{\circ}$ , so $∠ C O Y = ∠ Y O B = 6 5^{\circ}$ ; check $25 + 25 + 65 + 65 = 180$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a school wall clock is used to study angles. (i) Find the angle between the hands at 4:00. (ii) Find the angle between the hands at 3:40. (iii) Find the angle at 7:20. (iv) Is the angle at 3:40 a right angle? Give a reason.
Hint
Hour hand: $3 0^{\circ}$ per hour plus $0. 5^{\circ}$ per minute; minute hand: $6^{\circ}$ per minute.
Answer
(i) $4 \times 3 0^{\circ} = 12 0^{\circ}$ . (ii) Hour hand $= 9 0^{\circ} + 2 0^{\circ} = 11 0^{\circ}$ , minute hand $= 24 0^{\circ}$ , difference $13 0^{\circ}$ . (iii) Hour hand $= 21 0^{\circ} + 1 0^{\circ} = 22 0^{\circ}$ , minute hand $= 12 0^{\circ}$ , difference $10 0^{\circ}$ . (iv) No, $13 0^{\circ} \neq = 9 0^{\circ}$ ; the hour hand has moved past the 3.

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.Linear pair adds to?Show answer
$18 0^{\circ}$ .
2.Vertically opposite angles are?Show answer
Equal.
3.Angles round a point add to?Show answer
$36 0^{\circ}$ .
4.Complement of $3 5^{\circ}$ ?Show answer
$5 5^{\circ}$ .
5.Bisectors of a linear pair make what angle?Show answer
$9 0^{\circ}$ .

Linear Pairs, Vertically Opposite Angles and Angles at a Point: frequently asked questions

Are all adjacent angles a linear pair?

No. Adjacent angles form a linear pair only if their non-common arms are opposite rays, so that they add to 180^\circ.

Can an angle be both complementary and supplementary to the same angle?

No. The complement and the supplement of an angle differ by 90^\circ.

Why are vertically opposite angles equal?

Each is the supplement of the same adjacent angle, so they are equal.