Lines and Angles builds the angle facts that every later geometry chapter uses. It starts with rays, angles and their measures, then the pairs of angles formed when lines meet: complementary, supplementary, adjacent, linear pairs and vertically opposite angles. It proves the linear pair theorem and its converse, introduces proof by contradiction, and studies parallel lines cut by a transversal through corresponding, alternate and co-interior angles. It ends with the angle sum of a triangle and why a triangle must have at least two acute angles.
Key Concepts
1. Points, Lines, Line Segments and Rays
Geometry starts with three ideas we understand without a formal definition: a point (a position with no size), a line (straight, with no thickness, going on without end in both directions) and a plane (a flat surface going on without end). All figures in this chapter lie in one plane.
- A line AB passes through points A and B and extends endlessly both ways. It has no end points and cannot be measured. Lines are also named with a single small letter, such as line l.
- A line segment AB is the part of the line between A and B, including both. It has two end points and a definite length, written AB.
- A ray OA starts at the point O and goes on without end in one direction, through A. It has exactly one end point. The starting point is always written first, so ray OA and ray AO are different rays.
Collinear points. Three or more points that lie on one line are collinear; otherwise they are non-collinear. Exactly one line passes through two distinct points, so “A, O and B are collinear” means the same as “AOB is a straight line”.
Opposite rays. If O lies on line AB between A and B, rays OA and OB point in exactly opposite directions. They are opposite rays and together make up line AB.
Intersecting and parallel lines. Two distinct lines in a plane either meet at exactly one point (intersecting lines) or never meet, however far they are extended (parallel lines).
2. Angles and Their Measures
An angle is formed by two rays that have a common starting point. The common starting point is the vertex and the two rays are the arms of the angle. The angle formed by rays OA and OB is written ∠AOB or ∠BOA. The vertex always sits in the middle of the name.
The measure of an angle is the amount of turning needed to rotate one arm onto the other about the vertex. It is measured in degrees with a protractor. A full turn, bringing an arm back to where it started, is 360°.
Right angle and straight angle.
- A right angle is a quarter turn and measures 90°. Two lines that meet at a right angle are perpendicular, written l ⊥ m.
- A straight angle is formed when the two arms are opposite rays, so that they lie along one line pointing in opposite directions. It is a half turn and measures 180°.
- So a straight angle equals two right angles: 2 × 90° = 180°. A complete turn equals four right angles: 4 × 90° = 360°.
Classifying angles by measure.
| Type | Measure |
|---|---|
| Acute angle | More than 0° and less than 90° |
| Right angle | Exactly 90° |
| Obtuse angle | More than 90° and less than 180° |
| Straight angle | Exactly 180° |
| Reflex angle | More than 180° and less than 360° |
| Complete angle | Exactly 360° |
Reflex angle. Two rays OA and OB with a common vertex actually mark out two angles: the smaller one, ∠AOB, and the larger one on the other side, called reflex ∠AOB. Together they make a full turn, so
reflex ∠AOB = 360° - ∠AOB
For example, if ∠AOB = 125°, then reflex ∠AOB = 360° - 125° = 235°. When a question writes just ∠AOB, it means the smaller angle (less than 180°). The reflex angle is used only when it is named as reflex.
3. Pairs of Angles
Learn each definition word for word, because each has more than one condition.
Complementary angles. Two angles whose measures add to 90°. Each is the complement of the other. The complement of x is 90° - x. For example, 35° and 55° are complementary. Only acute angles have complements.
Supplementary angles. Two angles whose measures add to 180°. Each is the supplement of the other. The supplement of x is 180° - x. For example, 110° and 70° are supplementary.
Adjacent angles. Two angles are adjacent if all three conditions hold:
- they have a common vertex,
- they have a common arm, and
- their non-common arms lie on opposite sides of the common arm.
If ray OC lies inside ∠AOB, then ∠AOC and ∠COB are adjacent, with common vertex O and common arm OC. Their sum is ∠AOB: ∠AOC + ∠COB = ∠AOB. Two angles that share only a vertex, or that overlap, are not adjacent.
Linear pair. Two adjacent angles form a linear pair when their non-common arms are opposite rays, that is, when the non-common arms together form a straight line. Picture a straight line AB with a point O on it and a ray OC standing on the line, going upward from O. The two angles ∠AOC and ∠COB form a linear pair.
Vertically opposite angles. When two lines AB and CD intersect at O, four angles are formed around O. The angles facing each other across O, with no common arm, are vertically opposite. There are exactly two such pairs: ∠AOC with ∠BOD, and ∠AOD with ∠BOC.
Every linear pair is supplementary, but supplementary angles need not form a linear pair. Angles of 100° and 80° in two different corners of a page share no vertex, so they are not a linear pair.
4. The Linear Pair Theorem and Its Converse
Linear pair theorem. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180°.
Given: A line AB, a point O on it between A and B, and a ray OC not along the line.
To prove: ∠AOC + ∠COB = 180°.
Proof: Rays OA and OB are opposite rays, so ∠AOB is a straight angle and ∠AOB = 180°. Ray OC lies between OA and OB, so ∠AOC and ∠COB are adjacent and together make up ∠AOB. Hence ∠AOC + ∠COB = ∠AOB = 180°. ∴ The angles of a linear pair are supplementary.
Some books take this statement as a basic assumption and call it the Linear Pair Axiom. It is the tool behind almost every proof in this chapter.
Converse of the linear pair theorem. If the sum of two adjacent angles is 180°, then their non-common arms form a straight line.
Given: Adjacent angles ∠AOC and ∠COB with common arm OC and ∠AOC + ∠COB = 180°.
To prove: A, O and B are collinear, so AOB is a line.
Proof (by contradiction): Suppose OB is not the ray opposite to OA. Extend AO beyond O to get the opposite ray, and call it OB′; it lies on the same side of OC as OB. Now AOB′ is a line and OC stands on it, so by the linear pair theorem, ∠AOC + ∠COB′ = 180°. We are given ∠AOC + ∠COB = 180°. Comparing, ∠COB′ = ∠COB. Two rays OB and OB′ on the same side of OC making the same angle with OC must be the same ray. So OB and OB′ coincide, which contradicts our supposition. ∴ AOB is a straight line.
The converse is the test used to prove three points are collinear: show that two adjacent angles at the middle point add to 180°.
Consequences worth remembering.
- Angles around a point add to 360°. Extend any one ray backwards through the point to make a line; the angles on each side add to 180°, giving 180° + 180° = 360° in total.
- The bisectors of the two angles of a linear pair are perpendicular. If ∠AOC = x and ∠COB = y with x + y = 180°, the angle between the bisectors is x/2 + y/2 = (x + y)/2 = 90°.
5. Intersecting Lines and Vertically Opposite Angles
When two lines AB and CD cross at O, four angles are formed: ∠AOC, ∠COB, ∠BOD and ∠DOA. Next to each other they form linear pairs; across from each other they are vertically opposite.
Theorem. If two lines intersect each other, then the vertically opposite angles are equal.
Given: Lines AB and CD intersect at O.
To prove: ∠AOC = ∠BOD and ∠AOD = ∠BOC.
Proof:
- Ray OC stands on line AB. So ∠AOC + ∠COB = 180° (linear pair) … (1)
- Ray OB stands on line CD. So ∠COB + ∠BOD = 180° (linear pair) … (2)
- From (1) and (2), ∠AOC + ∠COB = ∠COB + ∠BOD.
- Subtracting ∠COB from both sides, ∠AOC = ∠BOD.
- In the same way, ray OA stands on line CD, so ∠AOD + ∠AOC = 180°; with (1), ∠AOD + ∠AOC = ∠AOC + ∠COB, so ∠AOD = ∠BOC.
What this means in practice. Once you know one of the four angles at a crossing, you know all four. If ∠AOC = 50°, then ∠BOD = 50° (vertically opposite), and ∠COB = ∠AOD = 180° - 50° = 130° (linear pairs).
Proof by contradiction
A proof by contradiction works in three steps:
- Assume the opposite of what you want to prove.
- Reason correctly from that assumption, using known facts.
- Reach a statement that cannot be true: it clashes with a known fact, with the given data, or with the assumption itself. So the assumption was false, and the original statement is true.
Example: two distinct lines meet in at most one point. Suppose two distinct lines l and m meet at two different points P and Q. Then both l and m pass through P and Q. But exactly one line passes through two distinct points, so l and m are the same line. This contradicts the fact that they are distinct. ∴ Two distinct lines have at most one common point.
The converse of the linear pair theorem (Concept 4) and the triangle facts in Concept 7 are also proved this way. Always write “Suppose…” at the start and “This contradicts…, so our supposition is false” at the end. Both sentences are part of the proof.
6. Parallel Lines and a Transversal
Two lines in the same plane are parallel if they do not meet, however far they are extended in either direction. We write l ∥ m. The distance between two parallel lines is the same everywhere, like the two rails of a straight railway track.
A transversal is a line that intersects two or more lines at distinct points. A transversal cutting two lines makes eight angles, four at each crossing.
Naming the eight angles
Picture two horizontal lines, l on top and m below, and a transversal t slanting down across them, cutting l at P and m at Q. At P, call the angles:
- ∠1 above l on the left of t, ∠2 above l on the right of t,
- ∠3 below l on the right of t, ∠4 below l on the left of t.
At Q, in the same positions: ∠5 above m on the left, ∠6 above m on the right, ∠7 below m on the right, ∠8 below m on the left.
| Type of angle or pair | Angles | How to spot it |
|---|---|---|
| Exterior angles | ∠1, ∠2, ∠7, ∠8 | Outside the strip between l and m |
| Interior angles | ∠3, ∠4, ∠5, ∠6 | Inside the strip between l and m |
| Corresponding angles (4 pairs) | ∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7, ∠4 and ∠8 | Same position at each crossing |
| Alternate interior angles (2 pairs) | ∠4 and ∠6, ∠3 and ∠5 | Between the lines, on opposite sides of t |
| Alternate exterior angles (2 pairs) | ∠1 and ∠7, ∠2 and ∠8 | Outside the lines, on opposite sides of t |
| Co-interior angles (2 pairs) | ∠4 and ∠5, ∠3 and ∠6 | Between the lines, on the same side of t |
Co-interior angles are also called interior angles on the same side of the transversal. These names describe positions and apply to any two lines. The equalities below hold only when the lines are parallel.
The corresponding angles property
If a transversal intersects two parallel lines, then each pair of corresponding angles is equal. Conversely, if a transversal intersects two lines so that a pair of corresponding angles is equal, then the two lines are parallel.
In the usual school treatment these two statements are taken as basic assumptions (the corresponding angles axiom and its converse), closely tied to Euclid’s fifth postulate. The other angle facts are proved from them.
Alternate interior angles are equal
Theorem. If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Proof: With l ∥ m and the labels above, ∠4 = ∠2 (vertically opposite angles at P) and ∠2 = ∠6 (corresponding angles, l ∥ m). So ∠4 = ∠6. In the same way, ∠3 = ∠1 (vertically opposite) and ∠1 = ∠5 (corresponding), so ∠3 = ∠5.
Co-interior angles are supplementary
Theorem. If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.
Proof: ∠4 = ∠8 (corresponding angles, l ∥ m). At Q, ∠5 and ∠8 form a linear pair (they share the left part of m, and their other arms are opposite parts of t), so ∠5 + ∠8 = 180°. Replacing ∠8 by ∠4, ∠4 + ∠5 = 180°. Similarly ∠3 + ∠6 = 180°.
The converses: tests for parallel lines
Each theorem has a true converse. If a transversal cuts two lines and any one of these holds, the lines are parallel:
- a pair of corresponding angles is equal, or
- a pair of alternate interior angles is equal, or
- a pair of co-interior angles adds to 180°.
Proof of the alternate-angle test: Suppose ∠4 = ∠6. Also ∠4 = ∠2 (vertically opposite). So ∠2 = ∠6, a pair of equal corresponding angles, and by the converse of the corresponding angles axiom, l ∥ m.
Proof of the co-interior test: Suppose ∠4 + ∠5 = 180°. Also ∠5 + ∠8 = 180° (linear pair). So ∠4 = ∠8, a pair of equal corresponding angles, and l ∥ m.
Transitivity of parallelism
Theorem. Lines which are parallel to the same line are parallel to each other.
Given: Distinct lines a, b and c with a ∥ b and c ∥ b.
To prove: a ∥ c.
Proof: Draw a transversal t cutting a, b and c. Let ∠x, ∠y and ∠z be the angles t makes with a, b and c in the same (corresponding) position. Since a ∥ b, ∠x = ∠y (corresponding angles). Since c ∥ b, ∠z = ∠y (corresponding angles). So ∠x = ∠z. These are corresponding angles made by t with a and c, and they are equal, so a ∥ c (converse of the corresponding angles axiom).
7. Angle Facts of a Triangle
Angle sum property. The sum of the angles of a triangle is 180°, two right angles.
Given: A triangle ABC.
To prove: ∠A + ∠B + ∠C = 180°.
Construction: Through A, draw a line PQ parallel to BC, with P on the same side of A as B and Q on the same side as C.
Proof:
- PQ ∥ BC and AB is a transversal, so ∠PAB = ∠ABC (alternate interior angles).
- PQ ∥ BC and AC is a transversal, so ∠QAC = ∠ACB (alternate interior angles).
- PAQ is a straight line, so the angles at A on one side of it add to 180°: ∠PAB + ∠BAC + ∠QAC = 180°.
- Substituting from steps 1 and 2: ∠ABC + ∠BAC + ∠ACB = 180°. ∴ ∠A + ∠B + ∠C = 180°.
Exterior angle property
If side BC of triangle ABC is extended beyond C to a point D, the angle ∠ACD is an exterior angle of the triangle. The two angles of the triangle at A and B, away from C, are its interior opposite angles.
An exterior angle of a triangle equals the sum of the two interior opposite angles: ∠ACD = ∠A + ∠B.
Proof: ∠ACD + ∠ACB = 180° (linear pair, since BCD is a line). Also ∠A + ∠B + ∠ACB = 180° (angle sum). Comparing, ∠ACD = ∠A + ∠B. It follows that an exterior angle is greater than each interior opposite angle.
What the angle sum rules out
A triangle cannot have two obtuse angles. Proof by contradiction: suppose ∠A > 90° and ∠B > 90°. Then ∠A + ∠B > 180°, and adding ∠C > 0° makes the total more than 180°. This contradicts the angle sum property. The same argument shows a triangle cannot have two right angles, or one right and one obtuse angle.
Every triangle has at least two acute angles. By the result above, at most one angle can be 90° or more. The other two must each be less than 90°, so they are acute.
A triangle cannot have all three angles less than 60°. Suppose each angle is less than 60°. Then the sum is less than 60° + 60° + 60° = 180°, which contradicts the angle sum property. So at least one angle is 60° or more. In the same way, all three angles cannot be more than 60°, so at least one angle is 60° or less. If all three are exactly 60°, the triangle is equiangular.
Classifying triangles by angles. A triangle is acute-angled (all angles acute), right-angled (one angle 90°) or obtuse-angled (one angle obtuse). In a right-angled triangle the two acute angles are complementary.
Formula and Theorem Sheet
| Result | Statement | How it is justified |
|---|---|---|
| Linear pair theorem | Ray OC stands on line AB: ∠AOC + ∠COB = 180° | The two angles make up a straight angle |
| Converse of linear pair | Adjacent angles add to 180°: the non-common arms form a line | Proof by contradiction |
| Angles around a point | Sum = 360° | Two straight angles |
| Vertically opposite angles | Lines AB, CD meet at O: ∠AOC = ∠BOD, ∠AOD = ∠BOC | Two linear pairs sharing an angle |
| Corresponding angles | l ∥ m: each pair equal; equal pair: l ∥ m | Axiom and its converse |
| Alternate interior angles | l ∥ m: each pair equal; and conversely | Corresponding + vertically opposite |
| Co-interior angles | l ∥ m: each pair sums to 180°; and conversely | Corresponding + linear pair |
| Transitivity | a ∥ b and b ∥ c give a ∥ c | Corresponding angles along one transversal |
| Angle sum of a triangle | ∠A + ∠B + ∠C = 180° | Parallel to BC through A, alternate angles |
| Exterior angle | Exterior angle = sum of the two interior opposite angles | Linear pair + angle sum |
| Triangle restrictions | At least two acute angles; no two obtuse angles; not all angles below 60° | Angle sum, by contradiction |
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Important Definitions
- Angle: the figure formed by two rays with a common starting point; the rays are the arms and the common point is the vertex.
- Complementary angles: two angles whose sum is 90°.
- Supplementary angles: two angles whose sum is 180°.
- Adjacent angles: two angles with a common vertex and a common arm, whose non-common arms lie on opposite sides of the common arm.
- Linear pair: two adjacent angles whose non-common arms are opposite rays.
- Vertically opposite angles: the pairs of non-adjacent angles formed at the point where two lines intersect.
- Perpendicular lines: two lines that intersect at a right angle.
- Parallel lines: two lines in the same plane that do not intersect, however far they are extended.
- Transversal: a line that intersects two or more lines at distinct points.
- Corresponding angles: angles in the same relative position at the two points where a transversal cuts two lines.
- Alternate interior angles: a pair of angles between the two lines, on opposite sides of the transversal, not adjacent to each other.
- Co-interior angles: a pair of angles between the two lines on the same side of the transversal.
Solved Examples
Example 1: An angle and its complement
Question: An angle is 24° more than its complement. Find the angle.
Solution: Let the angle be x. Its complement is 90° - x.
x = (90° - x) + 24°
2x = 114°
x = 57°
Check: the complement is 90° - 57° = 33°, and 57° - 33° = 24°. ∴ The angle is 57°.
Example 2: Supplement four times the complement
Question: The supplement of an angle is four times its complement. Find the angle.
Solution: Let the angle be x. Then
180° - x = 4(90° - x)
180° - x = 360° - 4x
3x = 180°
x = 60°
Check: supplement = 120°, complement = 30°, and 4 × 30° = 120°. ∴ The angle is 60°.
Example 3: A linear pair with algebra
Question: Ray OC stands on line AB. ∠AOC = (3x + 10)° and ∠COB = (2x - 5)°. Find x and both angles.
Solution: ∠AOC and ∠COB form a linear pair, so
(3x + 10) + (2x - 5) = 180
5x + 5 = 180
x = 35
∠AOC = 3(35) + 10 = 115° and ∠COB = 2(35) - 5 = 65°. Check: 115° + 65° = 180°.
Example 4: Using the converse to test collinearity
Question: Points A and B lie on opposite sides of a ray OC. (a) If ∠AOC = 108° and ∠COB = 72°, is AOB a straight line? (b) What if ∠AOC = 110° and ∠COB = 68°?
Solution: ∠AOC and ∠COB are adjacent (common vertex O, common arm OC, A and B on opposite sides of OC).
(a) 108° + 72° = 180°. By the converse of the linear pair theorem, the non-common arms OA and OB form a line. AOB is a straight line, so A, O and B are collinear.
(b) 110° + 68° = 178° ≠ 180°. So ∠AOB = 178°, which is not a straight angle, and A, O and B are not collinear.
Example 5: Vertically opposite angles and a reflex angle
Question: Lines AB and CD intersect at O. Ray OE lies inside ∠BOC. ∠AOC + ∠BOE = 75° and ∠BOD = 45°. Find ∠BOE and reflex ∠COE.
Solution:
- ∠AOC = ∠BOD = 45° (vertically opposite angles).
- ∠BOE = 75° - ∠AOC = 75° - 45° = 30°.
- Rays OC and OE stand on line AB on the same side, so ∠AOC + ∠COE + ∠EOB = 180°. So ∠COE = 180° - 45° - 30° = 105°.
- Reflex ∠COE = 360° - 105° = 255°.
Example 6: All four angles at a crossing
Question: Two lines intersect. One angle formed is (2x + 15)° and the angle vertically opposite to it is (3x - 20)°. Find all four angles.
Solution: Vertically opposite angles are equal:
2x + 15 = 3x - 20, so x = 35.
The angle is 2(35) + 15 = 85°, and its opposite is 3(35) - 20 = 85°. Each of the other two angles forms a linear pair with it: 180° - 85° = 95°.
∴ The four angles are 85°, 95°, 85° and 95°. Check: 85 + 95 + 85 + 95 = 360.
Example 7: All eight angles from one
Question: Lines l ∥ m are cut by transversal t. Using the labels of Concept 6 (∠1 to ∠4 at l, ∠5 to ∠8 at m), ∠1 = 65°. Find the other seven angles, giving a reason for each.
Solution:
| Angle | Value | Reason |
|---|---|---|
| ∠2 | 115° | Linear pair with ∠1 |
| ∠3 | 65° | Vertically opposite to ∠1 |
| ∠4 | 115° | Vertically opposite to ∠2 |
| ∠5 | 65° | Corresponding to ∠1 (l ∥ m) |
| ∠6 | 115° | Corresponding to ∠2, or alternate to ∠4 |
| ∠7 | 65° | Corresponding to ∠3 |
| ∠8 | 115° | Corresponding to ∠4 |
Example 8: Co-interior angles with algebra
Question: Lines l ∥ m are cut by a transversal. Two interior angles on the same side of the transversal are (2x + 20)° and (3x - 40)°. Find both angles.
Solution: Co-interior angles between parallel lines are supplementary:
(2x + 20) + (3x - 40) = 180
5x - 20 = 180
x = 40
The angles are 2(40) + 20 = 100° and 3(40) - 40 = 80°.
Example 9: Are the lines parallel?
Question: A transversal cuts lines p and q. In each case decide whether p ∥ q. (a) A pair of alternate interior angles are 58° and 58°. (b) A pair of co-interior angles are 105° and 70°.
Solution:
(a) Alternate interior angles are equal, so p ∥ q (converse of the alternate angles theorem).
(b) 105° + 70° = 175° ≠ 180°. So p is not parallel to q. The lines meet on the side of the transversal where these two angles lie, because their sum is less than 180°.
Example 10: The angle at a bend between parallel lines
Question: AB ∥ CD. Point E lies between the two lines. A and C are on the same side (the right side) of B and D, and E is also to the right of B and D. ∠ABE = 40° and ∠CDE = 35°. Find ∠BED.
Solution:
- Construction: through E, draw line EF parallel to AB, with F on the left side of E. Since AB ∥ CD, by transitivity EF ∥ CD too.
- EF ∥ AB and BE is a transversal. ∠BEF and ∠ABE are alternate interior angles, so ∠BEF = ∠ABE = 40°.
- EF ∥ CD and DE is a transversal. ∠FED and ∠CDE are alternate interior angles, so ∠FED = ∠CDE = 35°.
- Ray EF lies inside ∠BED, so ∠BED = ∠BEF + ∠FED = 40° + 35° = 75°.
The general result for this arrangement: ∠BED = ∠ABE + ∠CDE.
Example 11: Triangle angles in a ratio, and an exterior angle
Question: (a) The angles of a triangle are in the ratio 3 : 5 : 7. Find them and classify the triangle. (b) An exterior angle of a triangle is 110° and the two interior opposite angles are in the ratio 2 : 3. Find all three angles of the triangle.
Solution:
(a) Let the angles be 3k, 5k, 7k. Then 15k = 180°, so k = 12°. The angles are 36°, 60° and 84°. All three are acute, so the triangle is acute-angled.
(b) Let the interior opposite angles be 2k and 3k. By the exterior angle property, 2k + 3k = 110°, so k = 22°. These angles are 44° and 66°. The third angle forms a linear pair with the exterior angle: 180° - 110° = 70°. The angles are 44°, 66° and 70°. Check: 44 + 66 + 70 = 180.
Example 12: Bisectors of co-interior angles meet at a right angle
Question: AB ∥ CD, and a transversal cuts AB at P and CD at Q. B and D lie on the same side of the transversal. The bisectors of ∠BPQ and ∠DQP meet at R. Prove that ∠PRQ = 90°.
Solution:
- ∠BPQ and ∠DQP are interior angles on the same side of the transversal, and AB ∥ CD. So ∠BPQ + ∠DQP = 180°.
- PR bisects ∠BPQ, so ∠RPQ = ½∠BPQ. QR bisects ∠DQP, so ∠RQP = ½∠DQP.
- Adding: ∠RPQ + ∠RQP = ½(∠BPQ + ∠DQP) = ½ × 180° = 90°.
- In triangle PQR, ∠PRQ = 180° - (∠RPQ + ∠RQP) = 180° - 90° = 90°. Hence proved.
Competency-Based Questions (with answers)
1. Case-based: The level crossing
A straight road crosses a straight railway line. The two rails run east to west and are parallel: the first rail is the northern one, the second rail the southern one. On the north side of the first rail, the road makes an angle of 70° with it, measured on the east side of the road.
(a) What angle does the road make with the second rail in the same position (north of the rail, east of the road)? The road is a transversal to two parallel rails, and the two angles are corresponding angles. So the angle is 70°.
(b) Find the angle on the north side of the first rail, on the west side of the road. It forms a linear pair with the 70° angle: 180° - 70° = 110°.
2. Case-based: A pair of scissors
The two blades of a pair of scissors are straight and cross at the screw, O. When the handles are opened so that the angle between them is 38°, a student measures the angle between the cutting edges.
(a) What is the angle between the cutting edges? Name the property. The handles and blades lie along two straight lines crossing at O, so the angle between the blades is vertically opposite to the angle between the handles. It is 38°, by the vertically opposite angles theorem.
(b) Find the angle between one handle and the blade attached to the other handle. It forms a linear pair with the 38° angle: 180° - 38° = 142°.
(c) The handles are opened further until all four angles at O are equal. What is each angle and what is the relation between the two blades? Four equal angles adding to 360° give 90° each. The blades are then perpendicular.
3. Assertion-Reason
Options for Questions 3 to 5: (a) A and R true, R explains A; (b) A and R true, R does not explain A; (c) A true, R false; (d) A false, R true.
Assertion (A): Two angles measuring 130° and 50° always form a linear pair. Reason (R): The two angles of a linear pair are supplementary.
Answer: (d) A is false, but R is true. 130° and 50° are supplementary, but a linear pair also needs the angles to be adjacent with their non-common arms forming a line. Two separate angles with these measures need not even share a vertex.
4. Assertion-Reason
Assertion (A): If two lines intersect and one of the angles formed is 90°, then each of the other three angles is 90°. Reason (R): Vertically opposite angles are equal, and angles of a linear pair add to 180°.
Answer: (a) Both A and R are true, and R is the correct explanation of A. The angle opposite the 90° angle is 90°; each neighbour is 180° - 90° = 90°.
5. Assertion-Reason
Assertion (A): A triangle cannot have two obtuse angles. Reason (R): An exterior angle of a triangle is greater than each of its interior opposite angles.
Answer: (b) Both A and R are true, but R is not the correct explanation of A. A follows from the angle sum property: two angles each more than 90° would already add to more than 180°. R is a true fact about exterior angles, but it is not the reason for A.
6. Error analysis: Riya’s triangle
Riya says she has drawn a triangle whose angles measure 55°, 58° and 59°, and that “a triangle with all angles less than 60° is possible because my protractor shows it”.
(a) Find the sum of her three measured angles. 55 + 58 + 59 = 172°.
(b) Explain why her claim must be wrong. The angles of every triangle add to 180°. If all three angles were less than 60°, their sum would be less than 180°, a contradiction. So at least one angle must be 60° or more.
7. Reasoning: Light between two parallel mirrors
Two plane mirrors are placed parallel to each other, facing each other. A ray of light PQ strikes the first mirror at Q, reflects along QR to strike the second mirror at R, and reflects again along RS. At each mirror, the angle of incidence equals the angle of reflection, both measured from the normal (the line perpendicular to the mirror at that point).
(a) Why are the normals at Q and R parallel? Each normal is perpendicular to its own mirror. A line perpendicular to one of two parallel lines is perpendicular to the other too (corresponding angles, 90° each). So the second mirror makes equal corresponding angles of 90° with both normals, and the normals are parallel.
(b) Show that RS ∥ PQ. Let the normals be QM and RN. With QR as transversal to the parallel normals, ∠MQR = ∠NRQ (alternate angles). Incidence equals reflection at each mirror, so ∠PQR = 2∠MQR and ∠QRS = 2∠NRQ. Hence ∠PQR = ∠QRS. These are equal alternate angles made by transversal QR with PQ and RS, so PQ ∥ RS.
8. Competency MCQ: Which angles can form a triangle?
Question: Which set of angles can be the angles of a triangle? (i) 90°, 60°, 40° (ii) 100°, 95°, -15° (iii) 35°, 65°, 80° (iv) 45°, 45°, 45°
Answer: (iii). Its sum is 35 + 65 + 80 = 180° and every angle is positive. Set (i) adds to 190°. Set (ii) contains a negative angle and has two obtuse angles. Set (iv) adds to 135°, and all three angles are below 60°, which is impossible.
Important Questions for Exams
1-Mark Questions
- How many end points does a ray have? One, its starting point.
- Find the reflex angle of 142°. 360° - 142° = 218°.
- Find the supplement of the complement of 35°. Complement = 55°; supplement of 55° = 125°.
- Two lines intersect and one of the angles formed is 47°. Find the angle vertically opposite to it. 47°, since vertically opposite angles are equal.
- If a ∥ b and b ∥ c, what is the relation between a and c? a ∥ c (lines parallel to the same line are parallel to each other).
2-Mark Questions
- Two supplementary angles differ by 34°. Find the angles. x + y = 180° and x - y = 34°. Adding: 2x = 214°, so x = 107° and y = 73°.
- Prove that the bisectors of the angles of a linear pair are perpendicular. Let ray OC stand on line AB, with ∠AOC = x and ∠COB = y, so x + y = 180°. Let OP bisect ∠AOC and OQ bisect ∠COB. Then ∠POC = x/2 and ∠COQ = y/2, and OC lies between OP and OQ. So ∠POQ = x/2 + y/2 = (x + y)/2 = 90°. Hence OP ⊥ OQ.
- Lines l ∥ m are cut by a transversal. One of the interior angles is 4 times its co-interior partner. Find both. x + 4x = 180°, so x = 36°. The angles are 36° and 144°.
- Can a triangle have two right angles? Justify. No. Two right angles already add to 180°. The third angle must be more than 0°, so the sum would exceed 180°, contradicting the angle sum property.
3-Mark Questions
- Two lines intersect so that the sum of two vertically opposite angles is 150°. Find all four angles. Vertically opposite angles are equal, so each is 150° ÷ 2 = 75°. Each of the other two angles forms a linear pair with a 75° angle: 180° - 75° = 105°. The angles are 75°, 105°, 75°, 105°.
- Lines PQ and RS intersect at O. ∠POR : ∠ROQ = 5 : 7. Find all the angles at O. ∠POR and ∠ROQ form a linear pair: 5k + 7k = 180°, k = 15°. ∠POR = 75°, ∠ROQ = 105°. Vertically opposite: ∠QOS = ∠POR = 75° and ∠SOP = ∠ROQ = 105°.
- In triangle ABC, side BC is extended to D. ∠ACD = 128° and ∠B = 56°. Find ∠A and ∠ACB. Exterior angle property: ∠A + ∠B = ∠ACD, so ∠A = 128° - 56° = 72°. Linear pair: ∠ACB = 180° - 128° = 52°. Check: 72 + 56 + 52 = 180.
- A transversal cuts two lines so that a pair of alternate interior angles are equal. Prove that the lines are parallel. Let the transversal t cut l at P and m at Q, with alternate interior angles ∠4 (at P) and ∠6 (at Q) equal. ∠4 = ∠2 (vertically opposite at P). So ∠2 = ∠6. These are corresponding angles, and they are equal, so by the converse of the corresponding angles axiom, l ∥ m.
5-Mark Questions
- Prove that the sum of the angles of a triangle is 180°. Hence prove that an exterior angle of a triangle equals the sum of the two interior opposite angles. Let ABC be a triangle. Through A draw PQ ∥ BC, with P on the side of B and Q on the side of C. With AB as transversal, ∠PAB = ∠ABC (alternate interior angles). With AC as transversal, ∠QAC = ∠ACB (alternate interior angles). PAQ is a line, so ∠PAB + ∠BAC + ∠QAC = 180°. Substituting, ∠ABC + ∠BAC + ∠ACB = 180°. For the second part, extend BC to D. ∠ACD + ∠ACB = 180° (linear pair). Also ∠A + ∠B + ∠ACB = 180°. So ∠ACD = ∠A + ∠B.
- In triangle PQR, side QR is extended to S. The bisectors of ∠PQR and ∠PRS meet at T. Prove that ∠QTR = ½∠QPR. In triangle PQR, ∠PRS is an exterior angle, so ∠PRS = ∠QPR + ∠PQR. Halving: ½∠PRS = ½∠QPR + ½∠PQR, that is, ∠TRS = ½∠QPR + ∠TQR … (1), since RT and QT are the bisectors. In triangle TQR, ∠TRS is an exterior angle (QRS is a line), so ∠TRS = ∠TQR + ∠QTR … (2). From (1) and (2), ∠TQR + ∠QTR = ½∠QPR + ∠TQR, so ∠QTR = ½∠QPR. Hence proved.
Common Mistakes and Examiner Tips
- Calling any supplementary pair a linear pair. A linear pair must be adjacent with the non-common arms forming a line. Supplementary alone only means the sum is 180°.
- Using parallel-line equalities for lines that are not stated to be parallel. Corresponding and alternate angles are equal only when the lines are parallel. If the question does not give l ∥ m, you cannot assume it from how a sketch looks.
- Mixing up alternate and co-interior angles. Both lie between the lines. Alternate angles are on opposite sides of the transversal and are equal. Co-interior angles are on the same side and add to 180°. Check the side before writing the relation.
- Writing a theorem where its converse is needed. To find an angle, use “l ∥ m, so angles are equal”. To prove lines parallel, use “angles are equal, so l ∥ m” and name it as the converse.
- Missing reasons in proofs. Every step needs a reason in brackets: (linear pair), (vertically opposite angles), (alternate interior angles, AB ∥ CD). A step with no reason leaves the proof incomplete.
- Forgetting the reflex angle. When the question says reflex ∠COE, find ∠COE first and then subtract from 360°. Giving the smaller angle as the answer is wrong.
- Stopping after finding x. If the question asks for the angles, substitute x back into every expression and check the sum (180° or 360°).
- Skipping the construction in the angle sum proof. Write “Through A, draw PQ ∥ BC” as a separate line labelled Construction. Without the parallel line the alternate angles have nothing to stand on.
- A contradiction proof without the assumption stated. Begin with “Suppose, if possible, …” and end with “This contradicts …, so our supposition is false.” Both sentences are part of the proof.
- Trusting the sketch. A line that looks straight or two lines that look parallel may not be. Decide from the given angle measures, as in Example 4, where 178° looks straight but is not.
Quick Revision Points
- Angle: two rays with a common starting point (vertex); the vertex is the middle letter of the name.
- Right angle 90°; straight angle 180° = two right angles; complete angle 360°.
- Acute: 0° to 90°; obtuse: 90° to 180°; reflex: 180° to 360°; reflex ∠AOB = 360° - ∠AOB.
- Complementary: sum 90°. Supplementary: sum 180°.
- Adjacent angles: common vertex, common arm, other arms on opposite sides of the common arm.
- Linear pair: adjacent angles whose other arms are opposite rays; they add to 180°.
- Converse: adjacent angles adding to 180° have their other arms in a line; used to prove collinearity.
- Angles on one side of a line at a point: 180°. Angles around a point: 360°.
- Two intersecting lines: vertically opposite angles are equal; one angle fixes all four.
- Proof by contradiction: suppose the opposite, reason correctly, reach something impossible.
- Parallel lines: in one plane, never meet. Transversal: cuts two or more lines at distinct points, making 8 angles.
- If l ∥ m: corresponding angles equal, alternate interior angles equal, co-interior angles add to 180°.
- Any one of these conditions holding means the lines are parallel (converses).
- Transitivity: a ∥ b and b ∥ c give a ∥ c.
- Angle sum of a triangle is 180°; proof uses a line through one vertex parallel to the opposite side.
- Exterior angle = sum of the two interior opposite angles; it is larger than each of them.
- A triangle has at least two acute angles and cannot have two obtuse angles or two right angles.
- A triangle cannot have all three angles below 60° or all three above 60°.
Weightage and Exam Pattern
Class 9 has no board exam. Lines and Angles is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, the chapter is part of Unit IV, Geometry, which carries 25 marks out of the 80-mark theory paper and is allotted 69 teaching periods. The unit has five chapters: Introduction to Euclid’s Geometry, Lines and Angles, Triangles (Congruence Theorems), 4-gons (Quadrilaterals) and Circles. The course structure does not split the 25 marks between these chapters. Internal assessment carries a separate 20 marks.
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The CBSE question paper design weights questions by typology: about 54% remembering and understanding, 24% applying, and 22% analysing, evaluating and creating. Here that means definitions and classification, finding unknown angles, and proofs such as the tests for parallel lines and the contradiction arguments about triangles.
These angle facts return in the Class 10 board chapters on triangles and circles, where proofs begin with steps such as “alternate angles, since DE ∥ BC”.
Class 9 Maths – swipe through all 10 cards to understand the whole chapter.
What an angle is
An angle is formed by two rays with a common starting point, the vertex.
The vertex is always the middle letter: ∠AOB has vertex O.
- A line has no end points, a ray has one, a segment has two
- Right angle 90°, complete angle 360°
- Opposite rays together form a straight line
Acute to reflex
Angles are sorted by measure into acute, right, obtuse, straight, reflex and complete.
Plain ∠AOB always means the smaller angle, below 180°.
- Acute: between 0° and 90°
- Obtuse: between 90° and 180°
- Reflex: between 180° and 360°
Complementary, supplementary, adjacent
Pairs of angles are named by their sum or by how they sit next to each other.
Adjacent needs three things: common vertex, common arm, other arms on opposite sides.
- Complementary angles add to 90°
- Supplementary angles add to 180°
- Supplementary angles need not be adjacent
The linear pair theorem
If a ray stands on a line, the two adjacent angles formed add to 180°.
Converse: adjacent angles adding to 180° have their outer arms on one line.
- Use the converse to prove three points collinear
- Angles around a point add to 360°
- Bisectors of a linear pair are perpendicular
Vertically opposite angles
When two lines intersect, each pair of vertically opposite angles is equal.
Proof: two linear pairs share ∠COB, so the remaining angles are equal.
- One angle at a crossing fixes all four
- Neighbouring angles at a crossing add to 180°
- One right angle at a crossing makes all four right angles
Eight angles, named by position
A transversal cutting two lines makes four angles at each crossing.
Positions are named even when the lines are not parallel; equalities need parallel lines.
- Corresponding: same position at each crossing
- Alternate interior: between the lines, opposite sides of t
- Co-interior: between the lines, same side of t
Angle facts when l ∥ m
Parallel lines cut by a transversal give equal and supplementary pairs.
Each of the eight angles equals ∠1 or 180° − ∠1.
- Corresponding angles property is taken as an axiom
- Alternate: corresponding plus vertically opposite
- Co-interior: corresponding plus linear pair
Converses and transitivity
Any one angle condition from a transversal proves two lines parallel.
To find angles use the theorem; to prove parallel use the converse.
- Equal corresponding angles: lines parallel
- Equal alternate angles: lines parallel
- Co-interior angles adding to 180°: lines parallel
Angle sum and exterior angle
Draw a line through one vertex parallel to the opposite side and use alternate angles.
Exterior angle = sum of the two interior opposite angles.
- At least two angles of a triangle are acute
- No triangle has two obtuse or two right angles
- Not all three angles can be below 60°
Proof by contradiction
Assume the opposite, reason correctly, and reach something impossible.
Start with Suppose and end by naming what is contradicted.
- Two distinct lines meet in at most one point
- Converse of the linear pair theorem
- A triangle cannot have two obtuse angles
📝 Practice Lines and Angles - 10 questions
Concept-check questions · with answers & explanations
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Why C: The clock face is a full turn of 360° divided into 12 equal hour marks, so each hour mark is 360 divided by 12, that is 30°. At 4 o’clock the minute hand points at 12 and the hour hand at 4, which is 4 marks apart. The angle is 4 times 30°, that is 120°, an obtuse angle.
Why not A: 90° would need the hands 3 marks apart, as at 3 o’clock.
Why not B: 60° is only 2 marks apart, as at 2 o’clock.
Why not D: 150° corresponds to 5 marks apart, as at 5 o’clock.
Remember: Each hour mark on a clock is 30°, since 360° divided by 12 is 30°.
Why C: Since OC stands on line AB, angle AOC + angle BOC = 180° by the linear pair theorem. OD bisects angle AOC, so angle DOC is half of angle AOC. OE bisects angle BOC, so angle COE is half of angle BOC. Adding, angle DOE = angle DOC + angle COE = half of (angle AOC + angle BOC) = half of 180° = 90°.
Why not A: 45° is half of 90°; halving was applied once too often.
Why not B: 180° is the full linear pair, but the bisectors cut each half in two, so only half the total lies between them.
Why not D: Whatever angle AOC is, the halves always add to half of 180°, so the answer is fixed.
Remember: The bisectors of a linear pair always meet at 90°.
Why A: Step 1 (vertically opposite angles are equal): angle AOC and angle BOD are vertically opposite, so they are equal; each is 140 divided by 2, that is 70°. Step 2 (linear pair theorem): ray OA stands on line CD, so angle AOC + angle AOD = 180°, giving angle AOD = 180 minus 70 = 110°.
Why not B: 70° is angle AOC or angle BOD, not the adjacent angle AOD.
Why not C: 140° is the given sum of two angles, not the measure of any single angle at O.
Why not D: 40° is 180° minus 140°, subtracting the sum of two angles instead of one angle.
Remember: A sum of two vertically opposite angles is simply twice one of them.
Why A: Step 1 (co-interior angle theorem): the lines are parallel, so 2x + 30 + 3x – 20 = 180. Step 2 (solve): 5x + 10 = 180, so 5x = 170 and x = 34. Step 3 (evaluate both angles): 2(34) + 30 = 98° and 3(34) – 20 = 82°. Check: 98 + 82 = 180. The larger angle is 98°.
Why not B: 82° is the smaller of the two angles.
Why not C: 34 is the value of x, not an angle.
Why not D: 68 is 2x alone, forgetting to add 30.
Remember: Set co-interior expressions to 180, solve for x, then substitute back and compare.
Why B: Step 1 (express in one variable): let angle B = x, so angle A = 2x and angle C = 2x + x = 3x. Step 2 (angle sum of a triangle is 180°): x + 2x + 3x = 180, so 6x = 180 and x = 30. Step 3 (classify): the angles are 60°, 30° and 90°. One angle is exactly 90°, so the triangle is right-angled at C.
Why not A: Acute-angled needs all three angles below 90°, but angle C is 90°.
Why not C: Obtuse-angled needs an angle above 90°; the largest is exactly 90°.
Why not D: Equilateral needs all angles equal to 60°, but here they are 60°, 30° and 90°.
Remember: Put every angle in terms of one variable, use 180°, then look at the largest angle to classify.
Why D: Step 1 (straight angle = 180°): let the smaller angle be x°, so the larger is (x + 35)° and x + x + 35 = 180. Step 2 (solve the linear equation): 2x = 145, so x = 72.5. Step 3 (find the larger angle): x + 35 = 107.5°. Check: 72.5 + 107.5 = 180.
Why not A: 72.5° is the smaller angle, not the larger one asked for.
Why not B: 90° would make the two parts equal, but one is 35° more than the other.
Why not C: 145° is 2x, the value before dividing by 2.
Remember: A ray from the vertex of a straight angle always makes two angles that add to 180°.
Why A: Step 1 (set up the ratio): let angle BOC = x°, then angle AOC = 3x°. Step 2 (linear pair theorem): ray OC stands on line AB, so angle AOC + angle BOC = 180°, giving 3x + x = 180. Step 3 (solve): 4x = 180, so x = 45. Hence angle BOC = 45° and angle AOC = 135°.
Why not B: 135° is angle AOC, the larger angle, not angle BOC.
Why not C: 60° would come from 3x = 180, ignoring the smaller angle in the sum.
Why not D: 90° would make the two angles equal, contradicting the 3 to 1 relation.
Remember: Ratio plus linear pair: write both angles in terms of x and set the sum to 180.
Why B: Step 1 (linear pair theorem): ray OR stands on line PQ, so angle POR + angle ROQ = 180°. Step 2 (use the ratio): let the angles be 5k and 7k, so 12k = 180 and k = 15, giving angle POR = 75° and angle ROQ = 105°. Step 3 (vertically opposite angles are equal): angle POS is opposite angle ROQ, so angle POS = 105°.
Why not A: 75° is angle POR, which is adjacent to angle POS, not equal to it.
Why not C: 90° would mean the lines are perpendicular, but the 5 : 7 ratio shows they are not.
Why not D: 15° is the value of k, the common multiplier, not an angle.
Remember: Ratio pieces add to 180° along a line; then jump across the point using equality.
Why C: Step 1 (corresponding angle theorem): since l ∥ m, the corresponding angles are equal, so 3x + 20 = 5x – 10. Step 2 (solve): 20 + 10 = 5x – 3x, so 30 = 2x and x = 15. Each corresponding angle is then 3(15) + 20 = 65°, and 5(15) – 10 = 65°, matching as expected.
Why not A: 30 is 2x, the value before dividing by 2.
Why not B: 5 comes from the sign slip 20 – 10 = 2x.
Why not D: 21.25 comes from 3x + 20 + 5x – 10 = 180, treating corresponding angles as supplementary.
Remember: Corresponding and alternate angles between parallel lines are equal; only co-interior angles add to 180°.
Why C: Suppose each of the three angles is less than 60°. Adding them, the total would be less than 60 + 60 + 60 = 180°. But the angles of every triangle add to exactly 180°. This contradiction shows the supposition is false, so at least one angle must be 60° or more.
Why not A: A parallel line through a vertex can always be drawn; the construction does not depend on the sizes of the angles.
Why not B: Angles below 60° are all acute; two obtuse angles is a different impossibility, not the one that applies here.
Why not D: Only an equilateral triangle has all angles equal to 60°; other triangles have other angles that still add to 180°.
Remember: Three angles below 60° cannot reach 180°, so one angle is always at least 60°.
Chapter Navigation
Previous: Sequences and Progressions Class 9 Notes
Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Exploring Algebraic Identities Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
Explore More
Frequently Asked Questions
Supplementary angles are any two angles that add to 180°, wherever they are. A linear pair must also be adjacent: a common vertex, a common arm, and non-common arms that form a straight line. Every linear pair is supplementary, but two supplementary angles in different places are not a linear pair.
Let lines AB and CD meet at O. Ray OC stands on AB, so ∠AOC + ∠COB = 180°. Ray OB stands on CD, so ∠COB + ∠BOD = 180°. The two sums are equal, so subtracting ∠COB gives ∠AOC = ∠BOD. The same method with ray OA on CD gives ∠AOD = ∠BOC.
The converse says that if two adjacent angles add to 180°, their non-common arms form a straight line. It is used to prove that three points are collinear: show that the two adjacent angles at the middle point add to 180°. It is proved by contradiction, comparing the given ray with the ray opposite to the first arm.
Check any one pair of angles. If a pair of corresponding angles is equal, or a pair of alternate interior angles is equal, or a pair of co-interior angles adds to 180°, the lines are parallel. If any of these fails, for example co-interior angles adding to 175°, the lines are not parallel and meet on that side.
It means lines parallel to the same line are parallel to each other: if a ∥ b and b ∥ c, then a ∥ c. The proof draws one transversal across all three lines. The corresponding angles at a and b are equal, and those at b and c are equal, so those at a and c are equal, which makes a ∥ c.
The three angles of a triangle add to 180°. Two obtuse angles would each be more than 90°, so together they would already exceed 180° before the third angle is added. That contradicts the angle sum property. For the same reason, every triangle has at least two acute angles and cannot have all three angles below 60°.
You assume the opposite of the statement you want to prove, reason correctly from that assumption, and reach something impossible, such as a clash with a known theorem or with the given data. The assumption must then be false, so the statement is true. Always begin with Suppose and end by naming the contradiction.
Class 9 has no board exam, so the chapter is tested in school exams. It belongs to Unit IV, Geometry, which carries 25 of the 80 theory marks in the 2026-27 course structure, shared with four other geometry chapters. Expect definitions, finding unknown angles, tests for parallel lines and short proofs with a reason for every step.
