- Basic terms: a line extends infinitely in both directions; a line segment has two endpoints; a ray has one endpoint and goes on forever in one direction. An angle is formed when two rays share a common endpoint (the vertex).
- Types of angles: acute (0 to 90 exclusive), right (90), obtuse (90 to 180 exclusive), straight (180), reflex (180 to 360 exclusive), complete (360).
- Pairs of angles: complementary (sum 90 deg), supplementary (sum 180 deg), adjacent (common vertex and arm, no overlap), linear pair (adjacent and supplementary), vertically opposite (equal — key theorem).
- Parallel lines and transversal: corresponding angles equal (axiom); alternate interior angles equal (theorem); co-interior (same-side interior) angles supplementary.
- Triangle: angle sum = 180 deg (proved using parallel lines); exterior angle = sum of the two non-adjacent interior angles.
- Board weightage: ~6 marks/year — typically one theorem proof (3 marks) and one application or calculation problem (2 to 3 marks).
1. Basic terms and definitions
Before studying angles, we need crisp definitions of the building blocks.
- Point: has no dimension — it marks a location. Usually named with a capital letter like $P$ or $Q$.
- Line: a straight path extending infinitely in both directions. Written as line $AB$. Through two distinct points exactly one line passes.
- Line segment: the part of a line between two endpoints $A$ and $B$, written $AB$. It has a definite, measurable length.
- Ray: starts at a point (the initial point or endpoint) and goes infinitely in one direction. Starts at $A$, passes through $B$ and beyond.
- Collinear points: three or more points that lie on the same line. If they do not all lie on one line, they are non-collinear.
- Intersecting lines: two lines in the same plane that meet at exactly one point of intersection.
- Concurrent lines: three or more lines that all pass through a single common point.
- Parallel lines: two lines in the same plane that never meet, no matter how far they are extended. Written $l \parallel m$. The perpendicular distance between parallel lines is always the same.
Angle: formed by two rays sharing a common endpoint $B$ (the vertex). The rays are the arms of the angle. Written $\angle ABC$ or $\angle B$. Measured in degrees using a protractor.
Interior and exterior of an angle: the region between the two arms is the interior; the region outside is the exterior.
2. Types of angles
Every angle falls into exactly one of these six categories:
| Type | Measure | Example |
|---|---|---|
| Acute angle | Between $0°$ and $90°$ (exclusive) | $45°,\ 30°,\ 75°$ |
| Right angle | Exactly $90°$ | Corner of a square |
| Obtuse angle | Between $90°$ and $180°$ (exclusive) | $120°,\ 135°,\ 150°$ |
| Straight angle | Exactly $180°$ | A straight line |
| Reflex angle | Between $180°$ and $360°$ (exclusive) | $210°,\ 300°$ |
| Complete angle | Exactly $360°$ | Full rotation |
Zero angle: both arms coincide, measuring $0°$ — sometimes listed separately. Note: a right angle is neither acute nor obtuse.
(a) $89°$ — acute; (b) $90°$ — right; (c) $91°$ — obtuse; (d) $179°$ — obtuse; (e) $180°$ — straight; (f) $270°$ — reflex; (g) $360°$ — complete.
3. Pairs of angles
Special relationships between two angles come up constantly in proofs and problems.
3.1 Complementary angles
Two angles whose measures add up to $90°$. Each is the complement of the other.
Example: $30°$ and $60°$ are complementary; $45°$ and $45°$ are complementary. The complement of $x°$ is $(90 - x)°$.
3.2 Supplementary angles
Two angles whose measures add up to $180°$. Each is the supplement of the other.
Example: $110°$ and $70°$ are supplementary; $90°$ and $90°$ are supplementary. The supplement of $x°$ is $(180 - x)°$.
3.3 Adjacent angles
Two angles are adjacent if they satisfy all three conditions simultaneously:
- They have a common vertex.
- They have a common arm.
- Their non-common arms are on opposite sides of the common arm (interiors do not overlap).
Example: $\angle AOB$ and $\angle BOC$ are adjacent if $OB$ is the common arm, $O$ is the common vertex, and $OA$, $OC$ lie on opposite sides of $OB$.
3.4 Linear pair of angles
Two adjacent angles form a linear pair when their non-common arms together form a straight line. A linear pair is always supplementary.
Converse (Axiom 6.2): If two adjacent angles are supplementary, their non-common arms form a straight line.
So if ray $OC$ stands on line $AB$, then $\angle AOC + \angle BOC = 180°$.
3.5 Vertically opposite angles
When two lines intersect at point $O$, four angles form. Angles that are across from each other (not adjacent, sharing only the vertex) are vertically opposite. The word "vertical" here refers to the common vertex, not to up-down direction.
4. Theorem 6.1 — Vertically opposite angles are equal (proof)
Statement: If two lines intersect each other, then the vertically opposite angles are equal.
Given: Lines $AB$ and $CD$ intersect at point $O$, forming angles $\angle AOC$, $\angle BOC$, $\angle BOD$, and $\angle AOD$.
To prove: $\angle AOC = \angle BOD$ and $\angle AOD = \angle BOC$.
Proof:
- Ray $OA$ stands on line $CD$. So $\angle AOC + \angle AOD = 180°$ [Linear pair axiom, Axiom 6.1] ...(1)
- Ray $OD$ stands on line $AB$. So $\angle AOD + \angle BOD = 180°$ [Linear pair axiom, Axiom 6.1] ...(2)
- From (1) and (2): $\angle AOC + \angle AOD = \angle AOD + \angle BOD$
- Subtracting $\angle AOD$ from both sides: $\angle AOC = \angle BOD$ $\blacksquare$
Similarly, rays $OC$ and $OB$ on line $AB$ give: $\angle BOC + \angle AOC = 180°$ and $\angle AOC + \angle AOD = 180°$, leading to $\angle BOC = \angle AOD$.
Two lines intersect and one of the four angles is $35°$. Find all four angles.
Solution: Let $\angle AOC = 35°$.
- $\angle BOD = 35°$ (vertically opposite to $\angle AOC$)
- $\angle AOD = 180° - 35° = 145°$ (linear pair with $\angle AOC$)
- $\angle BOC = 145°$ (vertically opposite to $\angle AOD$)
The four angles are $\mathbf{35°, 145°, 35°, 145°}$. Notice they come in two equal pairs.
Two lines intersect. One angle is $(3x + 20)°$ and its vertically opposite angle is $(5x - 10)°$. Find $x$ and the angle.
Solution: Vertically opposite angles are equal: $3x + 20 = 5x - 10 \Rightarrow 30 = 2x \Rightarrow x = 15$.
Angle $= 3(15) + 20 = 65°$.
5. Parallel lines and a transversal — the 8 angles
A transversal is a line that intersects two or more lines at distinct points. When a transversal $t$ crosses two lines $l$ and $m$, it creates exactly 8 angles — 4 at each intersection.
Label the angles at the upper intersection (where $t$ meets $l$) as $\angle 1, \angle 2, \angle 3, \angle 4$ (going clockwise from the top-left region). At the lower intersection (where $t$ meets $m$): $\angle 5, \angle 6, \angle 7, \angle 8$ in the same fashion.
Named angle pairs — four types:
- Corresponding angles (F-shape): Same position at each intersection — one above $l$ and one above $m$ (or both below), on the same side of the transversal. Pairs: $(\angle 1, \angle 5),\ (\angle 2, \angle 6),\ (\angle 3, \angle 7),\ (\angle 4, \angle 8)$.
- Alternate interior angles (Z-shape): Both between the two lines (the interior region), on opposite sides of the transversal. Pairs: $(\angle 3, \angle 6),\ (\angle 4, \angle 5)$.
- Alternate exterior angles: Both outside the two lines, on opposite sides of the transversal. Pairs: $(\angle 1, \angle 8),\ (\angle 2, \angle 7)$.
- Co-interior angles (C-shape): Both between the two lines, on the same side of the transversal. Also called same-side interior or consecutive interior angles. Pairs: $(\angle 3, \angle 5),\ (\angle 4, \angle 6)$.
Memory rule: "F equals, Z equals, C supplements."
6. Corresponding angles axiom (Axiom 6.3 and 6.4)
This is taken as an axiom (accepted without proof — it is the foundational assumption for parallel lines in this chapter).
Axiom 6.4 (Converse): If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel.
So for $l \parallel m$ with transversal $t$, we get $\angle 1 = \angle 5,\ \angle 2 = \angle 6,\ \angle 3 = \angle 7,\ \angle 4 = \angle 8$. And if any one of these equalities holds, we can conclude $l \parallel m$.
$l \parallel m$, transversal $t$. One angle at the upper intersection is $70°$. Find all 8 angles.
Solution: Say $\angle 1 = 70°$. Then $\angle 3 = 70°$ (vert. opp.); $\angle 2 = \angle 4 = 110°$ (linear pair). At the lower intersection: $\angle 5 = 70°$ (corr. to $\angle 1$); $\angle 7 = 70°$ (vert. opp.); $\angle 6 = \angle 8 = 110°$. All 8 angles: $70°, 110°, 70°, 110°, 70°, 110°, 70°, 110°$.
7. Theorem 6.2 — Alternate interior angles are equal (proof)
Statement: If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Given: $AB \parallel CD$, transversal $PQ$ intersects $AB$ at $G$ and $CD$ at $H$. The alternate interior angle pair is $\angle AGH$ and $\angle GHD$.
To prove: $\angle AGH = \angle GHD$.
Proof:
- $\angle PGA = \angle GHD$ [Corresponding angles, since $AB \parallel CD$, Axiom 6.3] ...(1)
- $\angle PGA = \angle AGH$ [Vertically opposite angles at $G$] ...(2)
- From (1) and (2): $\angle AGH = \angle GHD$ $\blacksquare$
Theorem 6.3 (Converse): If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.
$AB$ and $CD$ are two lines. A transversal $PQ$ cuts them. $\angle APQ = 50°$ and $\angle CQP = 130°$. Are $AB$ and $CD$ parallel?
Solution: $\angle APQ + \angle CQP = 50° + 130° = 180°$. These are co-interior angles. Their sum is $180°$, so by the converse of Theorem 6.4, $AB \parallel CD$.
Alternatively: $\angle DQP = 180° - 130° = 50° = \angle APQ$. Alternate interior angles are equal, confirming $AB \parallel CD$ by Theorem 6.3.
Two parallel lines are cut by a transversal. One angle at the upper parallel is $\angle GEA = 126°$. Find all angles formed.
Solution: $\angle GEA = 126°$ (given). $\angle GEB = 180° - 126° = 54°$ (linear pair at $E$). $\angle AEH = 54°$ (vertically opposite to $\angle GEB$). $\angle BEH = 126°$ (vertically opposite to $\angle GEA$). At the lower parallel (point $H$): $\angle EHF = 126°$ (corresponding to $\angle GEA$, since $AB \parallel CD$). $\angle GHC = 126°$ (vertically opposite to $\angle EHF$). $\angle EHC = 54°$ (linear pair with $\angle EHF$). $\angle GHD = 54°$ (vertically opposite to $\angle EHC$).
8. Theorem 6.4 — Co-interior angles are supplementary (proof)
Statement: If a transversal intersects two parallel lines, then each pair of co-interior (same-side interior) angles is supplementary.
Given: $AB \parallel CD$, transversal $PQ$ at $G$ and $H$. Co-interior pair: $\angle BGH$ and $\angle GHD$.
To prove: $\angle BGH + \angle GHD = 180°$.
Proof:
- $\angle AGH = \angle GHD$ [Alternate interior angles, Theorem 6.2] ...(1)
- $\angle AGH + \angle BGH = 180°$ [Linear pair at $G$, ray $GA$ on line $PQ$... wait — $\angle AGH$ and $\angle BGH$ are on straight line $AB$] ...(2)
- Substituting (1) into (2): $\angle GHD + \angle BGH = 180°$ $\blacksquare$
Theorem 6.5 (Converse): If a transversal intersects two lines such that a pair of co-interior angles is supplementary, then the two lines are parallel.
$l \parallel m$, transversal $t$. One co-interior angle is $75°$. Find the other.
Solution: Co-interior angles sum to $180°$: other angle $= 180° - 75° = 105°$.
$AB \parallel CD \parallel EF$, transversal cuts all three. $y : z = 3 : 7$. Find $x$.
Solution: Let $y = 3k$, $z = 7k$. Since $CD \parallel EF$, co-interior angles sum: $y + z = 180° \Rightarrow 10k = 180° \Rightarrow k = 18°$. So $y = 54°$. Since $AB \parallel CD$, co-interior angles: $x + y = 180° \Rightarrow x = 180° - 54° = 126°$.
Two straight lines $PQ$ and $RS$ intersect at $O$. $\angle POR : \angle ROQ = 5 : 7$. Find all four angles.
Solution: Let $\angle POR = 5k$, $\angle ROQ = 7k$. Linear pair: $5k + 7k = 180° \Rightarrow k = 15°$.
$\angle POR = 75°$, $\angle ROQ = 105°$, $\angle QOS = 75°$ (vert. opp.), $\angle POS = 105°$ (vert. opp.).
9. Angle sum property of a triangle (Theorem 6.7, proof)
Statement: The sum of the three angles of a triangle is $180°$.
Given: $\triangle ABC$.
To prove: $\angle A + \angle B + \angle C = 180°$.
Construction: Through vertex $A$, draw line $PQ$ parallel to side $BC$, i.e., $PQ \parallel BC$.
Proof:
- $PQ \parallel BC$ and $AB$ is a transversal cutting them. $\therefore\ \angle PAB = \angle ABC$ [Alternate interior angles, Theorem 6.2] ...(1)
- $PQ \parallel BC$ and $AC$ is a transversal cutting them. $\therefore\ \angle QAC = \angle BCA$ [Alternate interior angles, Theorem 6.2] ...(2)
- $\angle PAB + \angle BAC + \angle QAC = 180°$ [Angles on a straight line $PAQ$ at point $A$] ...(3)
- Substituting (1) and (2) into (3): $\angle ABC + \angle BAC + \angle BCA = 180°$ i.e., $\angle B + \angle A + \angle C = 180°$ $\blacksquare$
Important corollaries:
- A triangle can have at most one right angle.
- A triangle can have at most one obtuse angle.
- In a right triangle, the two acute angles are complementary (sum $= 90°$).
- Each angle of an equilateral triangle $= 60°$.
Two angles of a triangle are $55°$ and $65°$. Find the third angle.
Solution: Third angle $= 180° - 55° - 65° = 60°$. The triangle has angles $55°, 65°, 60°$ — all acute, so it is an acute triangle.
Angles of a triangle are in the ratio $2:3:4$. Find each angle.
Solution: Let angles be $2x, 3x, 4x$. Sum $= 180°$: $9x = 180° \Rightarrow x = 20°$. Angles: $\mathbf{40°, 60°, 80°}$. (All acute — acute triangle.)
In $\triangle PQR$, $\angle P = 60°$ and $\angle Q = 65°$. Find $\angle R$.
Solution: $\angle R = 180° - 60° - 65° = 55°$.
10. Exterior angle theorem (Theorem 6.8, proof)
Statement: If a side of a triangle is produced, the exterior angle so formed equals the sum of the two non-adjacent (remote) interior angles.
Given: $\triangle ABC$. Side $BC$ is produced to point $D$, forming exterior angle $\angle ACD$.
To prove: $\angle ACD = \angle BAC + \angle ABC$, i.e., $\angle ACD = \angle A + \angle B$.
Proof:
- In $\triangle ABC$: $\angle A + \angle B + \angle ACB = 180°$ [Angle sum property] ...(1)
- $\angle ACB + \angle ACD = 180°$ [Linear pair, since $BD$ is a straight line] ...(2)
- From (1) and (2): $\angle A + \angle B + \angle ACB = \angle ACB + \angle ACD$
- Cancelling $\angle ACB$ from both sides: $\angle ACD = \angle A + \angle B$ $\blacksquare$
An exterior angle of a triangle is $110°$ and one of the interior opposite angles is $45°$. Find the other interior opposite angle.
Solution: Exterior angle $= \angle A + \angle B \Rightarrow 110° = 45° + \angle B \Rightarrow \angle B = 65°$.
In the figure, $\angle PRS$ is an exterior angle of $\triangle PQR$ at $R$. $\angle QPR = 35°$, $\angle PQR = 40°$. Find $\angle PRS$.
Solution: $\angle PRS = \angle QPR + \angle PQR = 35° + 40° = 75°$.
Two interior angles of a triangle are $50°$ and $70°$. Find the exterior angle at the third vertex.
Solution: Exterior angle $= 50° + 70° = 120°$. (Also: third interior angle $= 180° - 50° - 70° = 60°$; exterior $= 180° - 60° = 120°$. Both methods agree.)
11. Theorem 6.6 — Lines parallel to the same line are parallel to each other
Statement: If two lines are each parallel to a third line, then they are parallel to each other.
Proof sketch: Let $l \parallel n$ and $m \parallel n$. Draw a transversal $t$ cutting all three. Since $l \parallel n$, corresponding angles formed by $t$ at $l$ and $n$ are equal. Since $m \parallel n$, corresponding angles formed by $t$ at $m$ and $n$ are equal. By transitivity, corresponding angles at $l$ and $m$ (formed by $t$) are equal. Hence $l \parallel m$ by Axiom 6.4. $\blacksquare$
Why it matters: Parallelism is an equivalence relation on lines — reflexive (a line is parallel to itself), symmetric (if $l \parallel m$ then $m \parallel l$), and transitive (this theorem). This underpins geometry of polygons and coordinate geometry.
12. Summary — all parallel line results at a glance
| Angle pair | If $l \parallel m$, then... | Converse (implies $l \parallel m$) |
|---|---|---|
| Corresponding | Equal (Axiom 6.3) | Yes (Axiom 6.4) |
| Alt. interior | Equal (Thm 6.2) | Yes (Thm 6.3) |
| Alt. exterior | Equal (follows from Thm 6.2) | Yes |
| Co-interior | Supplementary (Thm 6.4) | Yes (Thm 6.5) |
| Vertically opp. | Always equal (Thm 6.1, any lines) | Not a parallel-line result |
| Linear pair | Always supplementary (Axiom 6.1, any lines) | Not a parallel-line result |
Common mistakes to avoid:
- Co-interior angles are NOT equal — they are supplementary. Confusing co-interior with alternate interior is a top error.
- Always justify steps in proofs — write [Corresponding angles, $l \parallel m$] or [Vertically opposite] after every equation. Missing reasons lose marks.
- Exterior angle theorem: the exterior angle equals the sum of the two remote interior angles, not all three (that would be $180°$).
- Linear pair check: before forming a linear pair, confirm the two angles share a ray and the remaining rays form a straight line.
- Construction in proof: when asked to prove angle sum of triangle, always state the construction explicitly — "Draw $PQ \parallel BC$ through $A$."
- $25°$
- $115°$
- $125°$
- $295°$
- $60°$
- $120°$
- $240°$
- $40°$
- $70°$
- $110°$
- $140°$
- $20°$
- $50°$
- $60°$
- $70°$
- $80°$
- $35°$
- $45°$
- $55°$
- $75°$
- Co-interior angles
- Linear pair angles
- Alternate interior angles
- Supplementary angles
- $20°$
- $70°$
- $80°$
- $160°$
- $34$
- $36$
- $38$
- $40$
- Acute-angled
- Obtuse-angled
- Right-angled
- Equilateral
- $12$
- $15$
- $18$
- $20$
$\angle EFD = \angle AEF = 55°$ [Alternate interior angles, $AB \parallel CD$, Theorem 6.2]
$\angle CFE = 180° - \angle EFD = 180° - 55° = 125°$ [Linear pair at $F$]
Proof:
Ray $OA$ on line $CD$: $\angle AOC + \angle AOD = 180°$ (linear pair) ...(1)
Ray $OD$ on line $AB$: $\angle AOD + \angle BOD = 180°$ (linear pair) ...(2)
From (1) and (2): $\angle AOC + \angle AOD = \angle AOD + \angle BOD$
Subtracting $\angle AOD$: $\angle AOC = \angle BOD$.
Similarly, $\angle AOD = \angle BOC$. $\blacksquare$
Angle sum property: $2x + 3x + 4x = 180° \Rightarrow 9x = 180° \Rightarrow x = 20°$.
Angles: $\mathbf{40°,\ 60°,\ 80°}$.
$\angle A = 180° - 45° - 65° = 70°$ (angle sum property).
By Exterior Angle Theorem: $\angle ACD = \angle A + \angle B = 70° + 45° = \mathbf{115°}$.
Linear pair: $x + 2x = 180° \Rightarrow 3x = 180° \Rightarrow x = 60°$.
$\angle POR = 60°$, $\angle ROQ = 120°$, $\angle QOS = 60°$ (vert. opp. to $\angle POR$), $\angle POS = 120°$ (vert. opp. to $\angle ROQ$).
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