Angles, Lines and Triangles · Study Notes · Edexcel IGCSE 4MA1 Higher

Foundation angle facts and reasons. Edexcel mark schemes credit specific reasons — memorise the language.

At a glance

Angles in a triangle: sum to $180°$ .
Angles on a straight line: sum to $180°$ .
Angles around a point: sum to $360°$ .
Vertically opposite angles: equal.
Parallel + transversal: corresponding (=), alternate (=), co-interior ( $180°$ ).
Exterior angle of triangle = sum of two opposite interior.
Always state the REASON.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.1 — Use angle facts to solve problems.
4.1 — Use parallel-line properties.
4.1 — Apply triangle angle properties.
4.1 — Use exterior angle theorem.

Basic angle facts

▼

Sum to $180°$ on lines and in triangles. $360°$ around a point.

Memorise these:

Fact	Sum
Angles in a TRIANGLE	$180°$
Angles on a STRAIGHT LINE	$180°$
Angles around a POINT	$360°$
Angles in a QUADRILATERAL	$360°$
Vertically OPPOSITE angles	equal

Special triangles:

Equilateral: all sides equal, all angles $60°$ .
Isosceles: two sides equal, two angles equal.
Right-angled: one angle is $90°$ .
Scalene: all sides and angles different.

Worked qualitative. Why do angles on a line sum to $180°$ ?

A 'line' makes a half-turn ( $180°$ ).
All angles fitting along it must add to $180°$ .

Edexcel tip. When stating an angle answer, ALWAYS give the reason: 'angles in a triangle sum to $180°$ ', 'angles on a straight line sum to $180°$ '.

Triangle: $180°$ .
Straight line: $180°$ .
Point: $360°$ .
Vertically opposite: equal.
Always state reason.

Parallel lines and transversals

▼

Three angle types: corresponding (F), alternate (Z), co-interior (C).

When two parallel lines are crossed by a transversal, FORMED ANGLES come in three special pairs:

Corresponding (F-shape).

In the same position relative to each line and the transversal.
EQUAL.

Alternate (Z-shape).

On opposite sides of the transversal, between the parallel lines.
EQUAL.

Co-interior / Allied (C-shape).

On the same side of the transversal, between the parallel lines.
SUM TO $180°$ .

F = corresponding (equal), Z = alternate (equal), C = co-interior (sum 180°).

Mnemonic: F = corresponding (same direction). Z = alternate (zigzag). C = co-interior (curve under).

Worked example. Parallel lines crossed by transversal. One angle is $65°$ .

Corresponding angle: $65°$ .
Alternate angle: $65°$ .
Co-interior angle: $180 - 65 = 115°$ .

Edexcel tip. State the angle TYPE in your reason: 'alternate angles' or 'co-interior angles'. Mark schemes deduct for using just 'parallel lines'.

Corresponding (F): equal.
Alternate (Z): equal.
Co-interior (C): sum to $180°$ .
Mnemonic: F, Z, C.

Triangle angles

▼

Sum to $180°$ . Exterior = sum of two opposite interior.

Sum of interior angles = $180°$ .

Exterior angle theorem:

Exterior angle = sum of the TWO OPPOSITE interior angles.

(Equivalently: exterior + adjacent interior = $180°$ .)

The exterior angle equals the sum of the two interior angles not next to it.

Example. Triangle with interior angles $50°, 60°, 70°$ . Exterior at $70°$ vertex:

Method 1: opposite interior = $50° + 60° = 110°$ .
Method 2: $180 - 70 = 110°$ .

Isosceles triangles.

Two equal sides → two equal angles.
The equal angles are OPPOSITE the equal sides.
The 'base' angles are equal; the 'apex' is between the equal sides.

Worked qualitative. A triangle has one angle $40°$ between the two equal sides (apex). Find the base angles.

Sum: $180°$ . So $40 + 2x = 180$ , $x = 70°$ .
Both base angles = $70°$ .

Edexcel tip. Always identify the type of triangle in your reasoning. 'Isosceles, so base angles are equal' is a key step.

Triangle sum: $180°$ .
Exterior = sum of opposite interior.
Isosceles: equal sides opposite equal angles.

Quick recap

Triangle: $180°$ .
Line: $180°$ . Point: $360°$ .
Parallel + transversal: F, Z, C angles.
Exterior of triangle = opposite two interior.
Always state the reason.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Triangle: $180°$ .
Line: $180°$ .
Point: $360°$ .
Corresponding/Alternate: equal.
Co-interior: $180°$ .
Exterior: opposite two interior.

How it’s examined

Angle facts on every Higher Tier paper (3-7 marks). Often combined with polygons or proof. Examiner reports flag (1) not stating reasons, (2) confusing alternate vs co-interior, (3) wrong angle pairs in isosceles.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on angles, lines and triangles, written exactly the way a tutor would explain them at the board.

1Find an angle in a triangle
Foundation• triangle
▼
Question
A triangle has angles $50°$ , $70°$ and $x$ . Find $x$ .
Step-by-step solution
1. Step 1
  Angles in a triangle sum to $180°$ .
2. Step 2
  $x = 180 - 50 - 70 = 60°$ .
Answer
$x = 60°$
Examiner tip
STATE the reason: 'angles in a triangle sum to $180°$ '. Mark schemes give M1 for the rule, A1 for the answer.
2Parallel lines and transversal
Higher• Adapted from 4MA1/1H May/Jun 2024 Q8• parallel lines
▼
Question
Two parallel lines are crossed by a transversal. One angle is $65°$ . Find the value of the corresponding angle, the alternate angle, and the co-interior angle.
Step-by-step solution
1. Step 1
  Corresponding angles are equal: $65°$ .
2. Step 2
  Alternate angles are equal: $65°$ .
3. Step 3
  Co-interior (allied) angles sum to $180°$ : $180 - 65 = 115°$ .
Answer
Corresponding: $65°$ . Alternate: $65°$ . Co-interior: $115°$ .
Examiner tip
Memorise: corresponding ='F-shape' equal. Alternate ='Z-shape' equal. Co-interior ='C-shape' sum to $180°$ .
3Exterior angle of a triangle
Foundation• exterior angle
▼
Question
A triangle has interior angles $50°, 60°$ and $70°$ . Find the exterior angle at the $70°$ vertex.
Step-by-step solution
1. Step 1
  Exterior angle = sum of two opposite interior angles.
2. Step 2
  $50° + 60° = 110°$ .
3. Step 3
  Or: exterior + interior = $180°$ . $180 - 70 = 110°$ . Same.
Answer
$110°$
4Isosceles triangle
Foundation• isosceles
▼
Question
An isosceles triangle has one angle of $40°$ at its apex. Find the two base angles.
Step-by-step solution
1. Step 1
  In an isosceles triangle, two angles are EQUAL (the base angles).
2. Step 2
  Sum of all three: $180°$ . $40 + 2x = 180$ , so $2x = 140$ , $x = 70°$ .
Answer
Both base angles = $70°$ .
5Writing the reason a mark scheme accepts [1 mark]
Getting started• Paper 1H/2H style — 1 mark• parallel lines, alternate angles, geometrical reasons
▼
Question
$AB$ and $CD$ are parallel straight lines. A straight line crosses $AB$ at $P$ and $CD$ at $Q$ . $A$ and $D$ are on opposite sides of $PQ$ .

Angle $APQ$ = 57°.

A student writes: "Angle $PQD$ = 57° because they are Z angles."

Write down the reason the student should have given. [1]
Step-by-step solution
1. Step 1
  Angles $APQ$ and $PQD$ lie between the parallel lines on opposite sides of the transversal $PQ$ , so they are alternate angles.
2. Step 2
  B1 Give the reason as a full statement: alternate angles are equal.
Answer
Alternate angles are equal.
Examiner tip
Examiner tip: A reason must state the fact, not just name the angles. "Alternate angles" on its own is weaker than "alternate angles are equal"; always write the whole sentence.

Common mistake: Using letter-shape names such as "Z angles" or "F angles". Pearson mark schemes do not accept them, so the reason mark is lost even when the angle is right.
6Three algebraic angles on a straight line [2 marks]
Getting started• Paper 1H/2H style — 2 marks• angles on a straight line, forming equations
▼
Question
Three angles lie next to each other on a straight line.

Their sizes are $(x + 15)°$ , $2x°$ and $(3x - 45)°$ .

Work out the value of $x$ . [2]
Step-by-step solution
1. Step 1
  M1 Angles on a straight line add up to 180°.
  $x + 15 + 2x + 3x - 45 = 180$
2. Step 2
  A1 Collect like terms and solve.
  $6x - 30 = 180 \quad \Rightarrow \quad 6x = 210 \quad \Rightarrow \quad x = 35$
Answer
$x = 35$
Examiner tip
Examiner tip: Check by substituting: $50 + 70 + 60 = 180$ .

Common mistake: Getting the constant wrong when collecting: $15 - 45 = -30$ , so the equation is $6x - 30 = 180$ , not $6x + 30 = 180$ (which gives $x = 25$ ).
7Triangle angles described in words [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• angles in a triangle, forming equations
▼
Question
In triangle $ABC$ , angle $A$ is 20° more than angle $B$ , and angle $C$ is twice angle $B$ .

Work out the size of each angle of the triangle. [3]
Step-by-step solution
1. Step 1
  M1 Let angle $B = x°$ . Then angle $A = (x + 20)°$ and angle $C = 2x°$ . Angles in a triangle add up to 180°.
  $x + 20 + x + 2x = 180$
2. Step 2
  A1 Solve.
  $4x + 20 = 180 \quad \Rightarrow \quad 4x = 160 \quad \Rightarrow \quad x = 40$
3. Step 3
  A1 Work out all three angles.
  $A = 60^{\circ}, \quad B = 40^{\circ}, \quad C = 80^{\circ}$
Answer
Angle $A$ = 60°, angle $B$ = 40°, angle $C$ = 80°
Examiner tip
Examiner tip: Let the letter stand for the angle that the others are described from. Here both $A$ and $C$ are described in terms of $B$ , so $B = x$ keeps the expressions simple.

Common mistake: Giving only $x = 40$ . The question asks for each angle, so all three must be stated for the final mark.
8A perpendicular from the right angle [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• right-angled triangles, angles in a triangle
▼
Question
Triangle $ABC$ has a right angle at $A$ .

$D$ is the point on $BC$ such that $AD$ is perpendicular to $BC$ .

Angle $ABC$ = 34°.

Work out the size of angle $CAD$ . [3]
Step-by-step solution
1. Step 1
  M1 In triangle $ABC$ , angles in a triangle add up to 180°.
  $\angle ACB = 180^{\circ} - 90^{\circ} - 34^{\circ} = 56^{\circ}$
2. Step 2
  M1 Triangle $ADC$ has a right angle at $D$ and angle $ACD$ = 56°.
  $\angle CAD = 180^{\circ} - 90^{\circ} - 56^{\circ}$
3. Step 3
  A1 Work it out.
  $\angle CAD = 34^{\circ}$
Answer
34°
Examiner tip
Examiner tip: Angle $CAD$ equals angle $ABC$ . The perpendicular splits the right-angled triangle into two smaller triangles with the same three angles as the original, which is worth remembering as a check.

Common mistake: Using 90° in the wrong place, for example taking angle $CAD$ to be $90 - 34 = 56°$ . That is angle $BAD$ , in the other small triangle.
9An exterior angle twice an interior angle [4 marks]
Stretch• Paper 1H/2H style — 4 marks• exterior angle, isosceles triangles, angle proof
▼
Question
$PQR$ is a triangle. The side $QR$ is extended to the point $S$ .

Angle $PRS$ is twice the size of angle $QPR$ .

Show that triangle $PQR$ is isosceles.

Give reasons for your working. [4]
Step-by-step solution
1. Step 1
  M1 Let angle $QPR = x$ , so angle $PRS = 2x$ . Angle $PRS$ is the exterior angle at $R$ , so it equals angle $QPR$ + angle $PQR$ .
  $2x = x + \angle PQR$
2. Step 2
  B1 Reason: the exterior angle of a triangle is equal to the sum of the interior opposite angles.
3. Step 3
  A1 Solve for angle $PQR$ .
  $\angle PQR = x = \angle QPR$
4. Step 4
  B1 Conclude: angles $QPR$ and $PQR$ are equal, so triangle $PQR$ is isosceles, with $QR = PR$ (the sides opposite the equal angles).
Answer
Angle $PQR$ = angle $QPR$ , so triangle $PQR$ is isosceles.
Examiner tip
Examiner tip: Letting the unknown angle be $x$ turns the words into an equation straight away. A proof with a single numerical example (say $x = 40$ ) is not enough; the argument has to work for every value.

Common mistake: Stopping at $\angle PQR = x$ without saying what it means. The last mark is for the conclusion that two angles are equal and so the triangle is isosceles.
10Three isosceles triangles around a point [5 marks]
Stretch• Paper 1H/2H style — 5 marks• isosceles triangles, angles at a point, angle chasing
▼
Question
$O$ is a point inside triangle $ABC$ . The lines $OA$ , $OB$ and $OC$ divide triangle $ABC$ into three triangles, $OAB$ , $OBC$ and $OCA$ .

$OA = OB = OC$ .

Angle $OAB$ = 35° and angle $OBC$ = 30°.

Work out the size of angle $OCA$ .

Give reasons for your working. [5]
Step-by-step solution
1. Step 1
  M1 Triangle $OAB$ is isosceles ( $OA = OB$ ), so angle $OBA$ = 35° (base angles of an isosceles triangle are equal). Angles in a triangle add up to 180°.
  $\angle AOB = 180^{\circ} - 2 \times 35^{\circ} = 110^{\circ}$
2. Step 2
  M1 In the same way, triangle $OBC$ is isosceles ( $OB = OC$ ), so angle $OCB$ = 30°.
  $\angle BOC = 180^{\circ} - 2 \times 30^{\circ} = 120^{\circ}$
3. Step 3
  M1 The three angles at $O$ add up to 360° (angles at a point).
  $\angle AOC = 360^{\circ} - 110^{\circ} - 120^{\circ} = 130^{\circ}$
4. Step 4
  A1 Triangle $OCA$ is isosceles ( $OC = OA$ ), so its base angles are equal.
  $\angle OCA = (180^{\circ} - 130^{\circ}) \div 2 = 25^{\circ}$
5. Step 5
  B1 Reasons stated: base angles of an isosceles triangle are equal; angles in a triangle add up to 180°; angles at a point add up to 360°.
Answer
25°
Examiner tip
Examiner tip: Check with the whole triangle: angle $A = 35 + 25 = 60$ , angle $B = 35 + 30 = 65$ , angle $C = 30 + 25 = 55$ , and $60 + 65 + 55 = 180$ .

Common mistake: Using 180° instead of 360° at $O$ . The three angles at $O$ go all the way round the point, not along a straight line.

Question 1
Paper 1H/2H style — 1 mark1 mark
Here are four angles.

48°, 90°, 137°, 214°

Write down the angle that is obtuse. [1]
Model answer
An obtuse angle is more than 90° and less than 180°.

Answer: 137°
Why this scores
Mark scheme:
- B1 for 137° cao
Why this scores full marks: The script uses the definition. 90° is a right angle (not obtuse) and 214° is reflex; both are common wrong choices.
Question 2
Paper 1H/2H style — 2 marks2 marks
$AB$ and $CD$ are parallel straight lines. The straight line $EF$ crosses $AB$ at $G$ and $CD$ at $H$ . $E$ is above $AB$ , $F$ is below $CD$ , and $B$ and $D$ are on the same side of $EF$ .

Angle $EGB$ = 67°.

(a) Work out the size of angle $CHF$ . [1]

(b) Give reasons for your answer. [1]
Model answer
(a) Angle $GHD$ = angle $EGB$ = 67°, and angle $CHF$ = angle $GHD$ .

Answer: 67°

(b) Corresponding angles are equal, and vertically opposite angles are equal.
Why this scores
Mark scheme:
- B1 for 67°
- B1 for corresponding angles are equal and vertically opposite angles are equal
Why this scores full marks: The script names the middle angle, $GHD$ , so both steps are visible, and each step has its reason in full. One reason alone does not justify a two-step answer.
Question 3
Paper 1H/2H style — 2 marks2 marks
In an isosceles triangle, the angle between the two equal sides is three times the size of each of the other two angles.

Work out the size of the largest angle of the triangle. [2]
Model answer
Let each base angle be $x°$ , so the angle between the equal sides is $3x°$ .

$x + x + 3x = 180$

$5x = 180, \qquad x = 36$

$3 \times 36 = 108$

Answer: 108°
Why this scores
Mark scheme:
- M1 for $x + x + 3x = 180$ oe
- A1 for 108° cao
Why this scores full marks: The script uses two base angles, not one, so the equation has $5x$ . Using $x + 3x = 180$ gives $x = 45$ and a wrong answer of 135°. It also answers the question asked: the largest angle, not $x$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
Two straight lines cross at a point, making four angles.

One of the four angles is four times the size of an angle next to it.

Work out the size of each of the four angles. [3]
Model answer
Two angles next to each other lie on a straight line, so they add up to 180°.

$x + 4x = 180, \qquad 5x = 180, \qquad x = 36$

$4 \times 36 = 144$

Vertically opposite angles are equal, so the other two angles are also 36° and 144°.

Answer: 36°, 144°, 36°, 144°
Why this scores
Mark scheme:
- M1 for $x + 4x = 180$ oe
- A1 for 36° and 144°
- B1 for all four angles, 36°, 144°, 36° and 144°
Why this scores full marks: The script uses the straight line (180°), not the full turn, for two neighbouring angles, and then uses vertically opposite angles to complete the set. Using 360° gives $x = 72$ and angles of 72° and 288°, which cannot sit next to each other on a straight line. Check: $36 + 144 + 36 + 144 = 360$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
The sizes of the three angles of a triangle are $(4x - 6)°$ , $(2x + 12)°$ and $(x + 6)°$ .

Show that the triangle is right-angled. [3]
Model answer
Angles in a triangle add up to 180°:

$4x - 6 + 2x + 12 + x + 6 = 180$

$7x + 12 = 180, \qquad 7x = 168, \qquad x = 24$

$4(24) - 6 = 90$

The angles are 90°, 60° and 30°. One angle is 90°, so the triangle is right-angled.
Why this scores
Mark scheme:
- M1 for $4x - 6 + 2x + 12 + x + 6 = 180$ oe
- A1 for $x = 24$
- A1 for showing one angle is 90° with a conclusion that the triangle is right-angled
Why this scores full marks: The script finds $x$ , substitutes into every expression and finishes with a sentence that answers the question. In a "show that", a correct value with no conclusion loses the final mark.
Question 6
Paper 1H/2H style — 4 marks4 marks
$PQR$ is a triangle. The straight line $XPY$ passes through $P$ and is parallel to $QR$ . $X$ is on the same side of $P$ as $Q$ , and $Y$ is on the same side of $P$ as $R$ .

Angle $XPQ = (3x + 4)°$ , angle $QPR = (x + 35)°$ and angle $PRQ = (2x + 21)°$ .

Work out the size of angle $QPR$ . [4]
Model answer
Angle $YPR$ = angle $PRQ$ = $(2x + 21)°$ (alternate angles are equal)

Angles on a straight line add up to 180°:

$3x + 4 + x + 35 + 2x + 21 = 180$

$6x + 60 = 180, \qquad x = 20$

$\angle QPR = 20 + 35 = 55$

Answer: 55°
Why this scores
Mark scheme:
- B1 for angle $YPR = (2x + 21)°$ with the reason alternate angles are equal
- M1 for $3x + 4 + x + 35 + 2x + 21 = 180$ oe
- A1 for $x = 20$
- A1 for 55° cao
Why this scores full marks: The script moves the angle at $R$ up to $P$ using alternate angles, so all three angles sit on the straight line $XPY$ . This is exactly why the angles in any triangle add up to 180°. The check: angle $PQR$ = angle $XPQ$ = 64°, and $64 + 55 + 61 = 180$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
The straight lines $AB$ and $CD$ cross at the point $E$ .

The line $EF$ is perpendicular to $AB$ . $C$ and $F$ are on the same side of $AB$ , and $C$ and $A$ are on the same side of $EF$ .

Angle $CEF = (x + 12)°$ and angle $BED = (4x + 3)°$ .

Work out the size of angle $AEC$ .

Give a reason for your answer. [4]
Model answer
$EF$ is perpendicular to $AB$ , so angle $AEF$ = 90°.

$\angle AEC = 90 - (x + 12) = 78 - x$

Angle $AEC$ = angle $BED$ (vertically opposite angles are equal):

$78 - x = 4x + 3, \qquad 75 = 5x, \qquad x = 15$

$\angle AEC = 78 - 15 = 63$

Answer: 63°
Why this scores
Mark scheme:
- M1 for angle $AEC = 90 - (x + 12)$ oe
- M1 for $90 - (x + 12) = 4x + 3$ oe
- A1 for 63° cao
- B1 for vertically opposite angles are equal
Why this scores full marks: The script uses the right angle to write angle $AEC$ in terms of $x$ before equating it to its vertically opposite angle, and it answers the question asked: angle $AEC$ , not $x$ . Check: $4(15) + 3 = 63$ and $27 + 63 = 90$ . A common error is to equate angle $CEF$ with angle $BED$ , which are not vertically opposite.
Question 8
Paper 1H/2H style — 4 marks4 marks
Triangle $ABC$ has a right angle at $C$ . Angle $ABC$ is twice the size of angle $BAC$ .

The bisector of angle $ABC$ meets $AC$ at $D$ .

Work out the size of angle $ADB$ . [4]
Model answer
Let angle $BAC = x°$ , so angle $ABC = 2x°$ .

$x + 2x + 90 = 180, \qquad 3x = 90, \qquad x = 30$

Angle $BAC$ = 30° and angle $ABC$ = 60°, so angle $ABD$ = 60 ÷ 2 = 30°.

In triangle $ABD$ :

$\angle ADB = 180 - 30 - 30 = 120$

Answer: 120°
Why this scores
Mark scheme:
- M1 for $x + 2x + 90 = 180$ oe
- A1 for angle $BAC = 30$ and angle $ABC = 60$
- M1 for angle $ABD = 30$ and $180 - 30 -$ their angle $ABD$
- A1 for 120° cao
Why this scores full marks: The script uses triangle $ABD$ , whose angles at $A$ and $B$ are now both known. An equally good route is angle $BDC = 180 - 90 - 30 = 60$ in triangle $BCD$ , then $180 - 60 = 120$ on the straight line $ADC$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
The straight lines $AD$ and $BC$ cross at the point $E$ , with $E$ between $A$ and $D$ and between $B$ and $C$ .

$AB$ is parallel to $CD$ and $EA = EB$ .

Angle $AEB = (2x + 8)°$ and angle $EDC = (3x - 10)°$ .

Work out the value of $x$ .

Give reasons for your working. [5]
Model answer
$EA = EB$ , so angle $EAB$ = angle $EBA$ (base angles of an isosceles triangle are equal).

Angles in a triangle add up to 180°:

$\angle EAB = \frac{180 - (2x + 8)}{2} = 86 - x$

Angle $EDC$ = angle $EAB$ (alternate angles are equal, since $AB$ is parallel to $CD$ ):

$3x - 10 = 86 - x, \qquad 4x = 96$

Answer: $x = 24$
Why this scores
Mark scheme:
- M1 for angle $EAB = \dfrac{180 - (2x + 8)}{2}$ oe
- M1 for $3x - 10 = 86 - x$ oe, equating angle $EDC$ with angle $EAB$ (or angle $EBA$ )
- A1 for $x = 24$ cao
- B2 for all three reasons: base angles of an isosceles triangle are equal; angles in a triangle add up to 180°; alternate angles are equal (B1 for two of these reasons)
Why this scores full marks: The script expresses a base angle of the isosceles triangle in terms of $x$ , then carries it across the parallel lines with alternate angles, stating each fact in the standard wording. Check: with $x = 24$ , angle $AEB$ = 56°, the base angles are 62° each ( $56 + 62 + 62 = 180$ ), and angle $EDC = 3(24) - 10 = 62$ °. Writing "Z angles" for the alternate-angle step would lose a reason mark.
Question 10
Paper 1H/2H style — 5 marks5 marks
$AB$ and $CD$ are parallel straight lines. A straight line crosses $AB$ at $P$ and $CD$ at $Q$ . $B$ and $D$ are on the same side of $PQ$ .

The bisector of angle $BPQ$ and the bisector of angle $PQD$ meet at the point $R$ .

Show that angle $PRQ$ = 90°.

Give reasons for your working. [5]
Model answer
Angles $BPQ$ and $PQD$ are co-interior angles, so $\angle BPQ + \angle PQD = 180$ (co-interior angles add up to 180°).

$PR$ and $QR$ are the bisectors, so

$\angle RPQ + \angle RQP = \tfrac{1}{2}\angle BPQ + \tfrac{1}{2}\angle PQD = \tfrac{1}{2}(180) = 90$

In triangle $PQR$ (angles in a triangle add up to 180°):

$\angle PRQ = 180 - 90 = 90$

So angle $PRQ$ = 90°.
Why this scores
Mark scheme:
- B1 for angle $BPQ$ + angle $PQD$ = 180 with the reason co-interior angles add up to 180°
- M1 for angle $RPQ = \tfrac{1}{2}$ angle $BPQ$ and angle $RQP = \tfrac{1}{2}$ angle $PQD$
- M1 for angle $RPQ$ + angle $RQP$ = 90
- M1 for $180 - 90$ in triangle $PQR$
- A1 for a complete argument concluding angle $PRQ$ = 90°, with angles in a triangle add up to 180° stated
Why this scores full marks: The script works with general angles, not a chosen example, so the result is proved for every transversal. Substituting one value such as angle $BPQ$ = 70° only checks a single case and scores at most the B mark.

4MA1 Angles, Lines and Triangles

At a glance

What you’ll learn

Basic angle facts

Parallel lines and transversals

Triangle angles

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Find an angle in a triangle

2Parallel lines and transversal

3Exterior angle of a triangle

4Isosceles triangle

5Writing the reason a mark scheme accepts [1 mark]

6Three algebraic angles on a straight line [2 marks]

7Triangle angles described in words [3 marks]

8A perpendicular from the right angle [3 marks]

9An exterior angle twice an interior angle [4 marks]

10Three isosceles triangles around a point [5 marks]

Model AnswersFull mark-scheme-style answers

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