Angles and Angle Reasoning

Angle reasoning is the backbone of the whole Geometry strand of OCR GCSE Mathematics (J560). Almost every geometry question — whether it is about parallel lines, polygons, or later the circle theorems — rests on a small set of angle facts that you combine, one step at a time, to reach an answer. The skill that earns marks is not just getting the number: it is giving a reason for your answer, naming the geometric fact at each stage. Recalling the facts is AO1; chaining them into a multi-step deduction is AO2; and applying them inside an unfamiliar problem is AO3. This lesson builds from angles on a line up to interior and exterior angles of polygons, always with the reason written down.

Key Vocabulary

Term	Meaning
Acute angle	An angle less than $90^{\circ}$90∘
Obtuse angle	An angle between $90^{\circ}$90∘ and $180^{\circ}$180∘
Reflex angle	An angle between $180^{\circ}$180∘ and $360^{\circ}$360∘
Transversal	A straight line that crosses two or more other lines
Interior angle	An angle inside a polygon at a vertex
Exterior angle	The angle between a side and the extension of the next side
Polygon	A closed 2D shape with straight sides
Regular polygon	A polygon with all sides and all angles equal

Angle Facts on Lines and at a Point

Three facts cover almost every "find the missing angle" question. Angles on a straight line add up to $180^{\circ}$180∘, because a straight line represents a half-turn. Angles around a point add up to $360^{\circ}$360∘, a full turn. When two straight lines cross, vertically opposite angles are equal — the two angles "facing" each other across the crossing point are always the same size. None of these is on the J560 formula sheet, so you must know them and quote them by name; the reasoning marks in this topic are awarded for writing the fact down, not just the number.

It helps to picture an angle as an amount of turning rather than as a static corner. A right angle is a quarter-turn ($90^{\circ}$90∘), a straight line is a half-turn ($180^{\circ}$180∘), and a complete revolution is a full turn ($360^{\circ}$360∘). Every fact in this lesson is really a statement about how much turning fits into a particular configuration, which is why the numbers $90$90, $180$180 and $360$360 appear again and again.

The diagram below shows two straight lines meeting at a point, splitting the angles around it.

In this figure, $a$a and $c$c are vertically opposite, so $a = c$a=c; likewise $b = d$b=d. Also $a + b = 180^{\circ}$a+b=180∘ because they sit on a straight line.

Worked Example 1

Two angles on a straight line are $x$x and $124^{\circ}$124∘. Work out $x$x.

Angles on a straight line sum to $180^{\circ}$180∘, so $x = 180^{\circ} - 124^{\circ} = 56^{\circ}$x=180∘−124∘=56∘.

Answer: $x = 56^{\circ}$x=56∘.

Common error: writing $360^{\circ} - 124^{\circ}$360∘−124∘ by confusing "on a line" with "at a point".

Worked Example 2

Three angles meet at a point: $140^{\circ}$140∘, $95^{\circ}$95∘ and $y$y. Find $y$y.

Angles at a point sum to $360^{\circ}$360∘, so $y = 360^{\circ} - 140^{\circ} - 95^{\circ} = 125^{\circ}$y=360∘−140∘−95∘=125∘.

Answer: $y = 125^{\circ}$y=125∘.

Worked Example 3

Two straight lines cross. One angle is $72^{\circ}$72∘. Write down the sizes of the other three angles, giving a reason.

The vertically opposite angle is $72^{\circ}$72∘ (vertically opposite angles are equal). Each of the remaining two angles is $180^{\circ} - 72^{\circ} = 108^{\circ}$180∘−72∘=108∘ (angles on a straight line sum to $180^{\circ}$180∘).

Answer: $72^{\circ}$72∘, $108^{\circ}$108∘, $108^{\circ}$108∘.

Worked Example 3b

Around a point, four angles are $x$x, $2x$2x, $90^{\circ}$90∘ and $70^{\circ}$70∘. Work out $x$x.

Angles at a point sum to $360^{\circ}$360∘, so $x + 2x + 90^{\circ} + 70^{\circ} = 360^{\circ}$x+2x+90∘+70∘=360∘. That gives $3x + 160^{\circ} = 360^{\circ}$3x+160∘=360∘, so $3x = 200^{\circ}$3x=200∘ and $x = \dfrac{200^{\circ}}{3} \approx 66.7^{\circ}$x=3200∘​≈66.7∘.

Answer: $x \approx 66.7^{\circ}$x≈66.7∘ (or exactly $\dfrac{200}{3}\,^{\circ}$3200​∘).

Common error: forgetting one of the unknown terms — here it is easy to drop the $2x$2x and divide by the wrong amount.

Angles in Parallel Lines

When a transversal crosses two parallel lines (marked with matching arrows), three relationships appear. Alternate angles (a Z-shape) are equal. Corresponding angles (an F-shape) are equal. Co-interior angles, sometimes called allied angles (a C- or U-shape), add up to $180^{\circ}$180∘. The trick in the exam is to look at the shape the angles make with the transversal and the parallel lines: a Z means equal alternate angles, an F means equal corresponding angles, and a C or U means co-interior angles that sum to a straight line. Each of these must be quoted by name to earn the reasoning mark. The figure shows a transversal cutting two parallel lines.

Here $p$p and $q$q form a Z-shape across the parallel lines, so they are alternate angles and $p = q$p=q.

Worked Example 4

A transversal crosses two parallel lines. One alternate angle is $64^{\circ}$64∘. Work out the co-interior angle on the same side of the transversal, giving a reason.

The other alternate angle is $64^{\circ}$64∘ (alternate angles are equal). The co-interior angle pairs with it on a straight line, so it is $180^{\circ} - 64^{\circ} = 116^{\circ}$180∘−64∘=116∘ (co-interior angles sum to $180^{\circ}$180∘).

Answer: $116^{\circ}$116∘.

Worked Example 5

Lines $AB$AB and $CD$CD are parallel. A transversal makes a corresponding angle of $113^{\circ}$113∘ with $AB$AB. Find the equal corresponding angle at $CD$CD.

Corresponding angles are equal, so the angle at $CD$CD is $113^{\circ}$113∘.

Answer: $113^{\circ}$113∘.

Common error: treating corresponding angles as if they add to $180^{\circ}$180∘. Only co-interior angles do that.

Worked Example 6

In a diagram, $AB$AB is parallel to $CD$CD. A transversal gives angles $(3x + 20)^{\circ}$(3x+20)∘ and $(5x - 40)^{\circ}$(5x−40)∘ as co-interior angles. Find $x$x.

Co-interior angles sum to $180^{\circ}$180∘: $(3x + 20) + (5x - 40) = 180$(3x+20)+(5x−40)=180, so $8x - 20 = 180$8x−20=180, giving $8x = 200$8x=200 and $x = 25$x=25.

Answer: $x = 25$x=25.

Angles in Triangles and Quadrilaterals

The angles in any triangle add up to $180^{\circ}$180∘; in any quadrilateral they add up to $360^{\circ}$360∘. You can see why a quadrilateral totals $360^{\circ}$360∘: a diagonal splits it into two triangles, and $2 \times 180^{\circ} = 360^{\circ}$2×180∘=360∘. An exterior angle of a triangle equals the sum of the two opposite interior angles — this is a shortcut worth memorising because it saves a step. In an isosceles triangle the two base angles (the angles opposite the equal sides) are equal; an equilateral triangle has all three angles equal to $60^{\circ}$60∘. Recognising which type of triangle you are looking at is often the key that unlocks a multi-step problem.

Worked Example 7

An isosceles triangle has an angle of $40^{\circ}$40∘ between its two equal sides. Work out each base angle.

The two base angles are equal; call each $b$b. Then $40^{\circ} + 2b = 180^{\circ}$40∘+2b=180∘, so $2b = 140^{\circ}$2b=140∘ and $b = 70^{\circ}$b=70∘.

Answer: each base angle is $70^{\circ}$70∘.

Worked Example 8

In $\triangle PQR$△PQR, $\angle P = 55^{\circ}$∠P=55∘ and $\angle Q = 70^{\circ}$∠Q=70∘. Side $QR$QR is extended to $S$S. Work out the exterior angle $\angle PRS$∠PRS, giving a reason.

The exterior angle equals the sum of the two opposite interior angles: $\angle PRS = 55^{\circ} + 70^{\circ} = 125^{\circ}$∠PRS=55∘+70∘=125∘.

Answer: $125^{\circ}$125∘. (Check: interior $\angle PRQ = 180^{\circ} - 55^{\circ} - 70^{\circ} = 55^{\circ}$∠PRQ=180∘−55∘−70∘=55∘, and $55^{\circ} + 125^{\circ} = 180^{\circ}$55∘+125∘=180∘ on the straight line.)

Worked Example 9

Three angles of a quadrilateral are $90^{\circ}$90∘, $85^{\circ}$85∘ and $110^{\circ}$110∘. Work out the fourth angle.

The angles sum to $360^{\circ}$360∘: fourth angle $= 360^{\circ} - 90^{\circ} - 85^{\circ} - 110^{\circ} = 75^{\circ}$=360∘−90∘−85∘−110∘=75∘.

Answer: $75^{\circ}$75∘.

Interior and Exterior Angles of Polygons

For a polygon with $n$n sides, the sum of interior angles is $(n - 2) \times 180^{\circ}$(n−2)×180∘. The reason is the same diagonal idea: any $n$n-sided polygon can be cut into $(n - 2)$(n−2) triangles from a single vertex, and each triangle contributes $180^{\circ}$180∘. This formula holds for regular and irregular polygons alike — it depends only on the number of sides, not on whether the shape is symmetrical. The exterior angles of any convex polygon always add up to $360^{\circ}$360∘, however many sides there are, because walking once around the outside turns you through one full revolution. For a regular polygon the exterior angles are all equal, so each one is $\dfrac{360^{\circ}}{n}$n360∘​, and each interior angle is $180^{\circ}$180∘ minus that. Because dividing by $n$n is so quick, the exterior-angle route is usually the fastest way into a regular-polygon question. The figure shows a regular pentagon with one interior angle marked.

Worked Example 10

Work out the sum of the interior angles of a heptagon ($7$7 sides).

Sum $= (n - 2) \times 180^{\circ} = (7 - 2) \times 180^{\circ} = 5 \times 180^{\circ} = 900^{\circ}$=(n−2)×180∘=(7−2)×180∘=5×180∘=900∘.

Answer: $900^{\circ}$900∘.

Worked Example 11

Find the size of each interior angle of a regular octagon.

Sum $= (8 - 2) \times 180^{\circ} = 1080^{\circ}$=(8−2)×180∘=1080∘. Each interior angle $= 1080^{\circ} \div 8 = 135^{\circ}$=1080∘÷8=135∘.

Answer: $135^{\circ}$135∘. (Check via exterior angle: $\dfrac{360^{\circ}}{8} = 45^{\circ}$8360∘​=45∘, and $180^{\circ} - 45^{\circ} = 135^{\circ}$180∘−45∘=135∘.)

Worked Example 12

Each interior angle of a regular polygon is $156^{\circ}$156∘. Work out the number of sides.

Each exterior angle $= 180^{\circ} - 156^{\circ} = 24^{\circ}$=180∘−156∘=24∘. Then $n = \dfrac{360^{\circ}}{24^{\circ}} = 15$n=24∘360∘​=15.

Answer: $15$15 sides.

Common error: using $(n-2)\times 180 = 156$(n−2)×180=156 directly. The $156^{\circ}$156∘ is one angle, not the whole sum; go via the exterior angle instead.

Worked Example 13

A pentagon has four interior angles of $100^{\circ}$100∘, $115^{\circ}$115∘, $130^{\circ}$130∘ and $95^{\circ}$95∘. Work out the fifth interior angle.

The interior angles of a pentagon sum to $(5 - 2) \times 180^{\circ} = 540^{\circ}$(5−2)×180∘=540∘. The fifth angle $= 540^{\circ} - (100^{\circ} + 115^{\circ} + 130^{\circ} + 95^{\circ}) = 540^{\circ} - 440^{\circ} = 100^{\circ}$=540∘−(100∘+115∘+130∘+95∘)=540∘−440∘=100∘.

Answer: $100^{\circ}$100∘.

Worked Example 14

Each exterior angle of a regular polygon is $40^{\circ}$40∘. Work out how many sides it has, and the size of each interior angle.

Number of sides $= \dfrac{360^{\circ}}{40^{\circ}} = 9$=40∘360∘​=9, so it is a nonagon. Each interior angle $= 180^{\circ} - 40^{\circ} = 140^{\circ}$=180∘−40∘=140∘.

Answer: $9$9 sides; each interior angle $140^{\circ}$140∘.

Building a Chain of Reasons

The hardest angle questions on an OCR paper are not the ones with awkward numbers — they are the ones that need several facts joined together. The technique is always the same: find any angle you can work out from the diagram, write it down with its reason, and use it as a stepping stone to the next one. Never try to leap straight to the final answer. Annotating the diagram as you go, filling in each angle the moment you know it, turns a daunting problem into a sequence of single, easy steps. Examiners reward this method generously, because the working itself carries the marks even if the final number slips.

A good mental checklist when you are stuck is: are there any straight lines (so angles sum to $180^{\circ}$180∘)? Any crossing lines (vertically opposite angles)? Any parallel lines marked with arrows (alternate, corresponding or co-interior)? Any triangles or isosceles triangles hiding in the figure? Running through this list almost always reveals the next step.

Worked Example 15

In a diagram, a triangle $ABC$ABC sits on a straight line $BCD$BCD. $\angle ABC = 50^{\circ}$∠ABC=50∘ and $\angle BAC = 60^{\circ}$∠BAC=60∘. A line $CE$CE is drawn parallel to $BA$BA. Work out $\angle ECD$∠ECD, giving a reason for each step.

Step 1: In $\triangle ABC$△ABC, the angles sum to $180^{\circ}$180∘, so $\angle ACB = 180^{\circ} - 50^{\circ} - 60^{\circ} = 70^{\circ}$∠ACB=180∘−50∘−60∘=70∘ (angles in a triangle sum to $180^{\circ}$180∘).

Step 2: $CE$CE is parallel to $BA$BA, and $AC$AC is a transversal, so $\angle ECA = \angle BAC = 60^{\circ}$∠ECA=∠BAC=60∘ (alternate angles are equal).

Step 3: $\angle ACB$∠ACB, $\angle ECA$∠ECA and $\angle ECD$∠ECD lie on the straight line $BCD$BCD, so $\angle ECD = 180^{\circ} - 70^{\circ} - 60^{\circ} = 50^{\circ}$∠ECD=180∘−70∘−60∘=50∘ (angles on a straight line sum to $180^{\circ}$180∘).

Answer: $\angle ECD = 50^{\circ}$∠ECD=50∘.

Common error: picking the wrong pair of parallel-line angles in Step 2 — checking the Z-shape carefully confirms alternate angles are the ones being used.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: "$AB$AB is parallel to $CD$CD. A transversal meets $AB$AB at $X$X and $CD$CD at $Y$Y. The angle on the upper-right at $X$X is $74^{\circ}$74∘. Work out the angle on the lower-left at $Y$Y. Give a reason for your answer."

Grades 3–4 response: Spots that $74^{\circ}$74∘ and the required angle look equal and writes $74^{\circ}$74∘, but the reason is vague — "they are the same" — without naming alternate angles, so the reasoning mark is at risk.

Grades 5–6 response: States the angle is $74^{\circ}$74∘ because alternate angles are equal, identifying the Z-shape made by the transversal across the two parallel lines.

Grades 7–9 response: Gives $74^{\circ}$74∘, names alternate angles, and adds a confirming step — the co-interior angle would be $180^{\circ} - 74^{\circ} = 106^{\circ}$180∘−74∘=106∘ — to show the configuration is consistent, communicating the full reasoning chain.

Examiner-style commentary: the accuracy mark is the number; the reasoning mark needs the named angle fact, which weaker answers omit.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: "Each interior angle of a regular polygon is $162^{\circ}$162∘. Work out how many sides the polygon has. Show that your answer is a whole number."

Grades 3–4 response: Tries $(n-2)\times 180 = 162$(n−2)×180=162 and gets stuck, or guesses; may reach an answer by trial but cannot justify it.

Grades 5–6 response: Uses the exterior angle: $180^{\circ} - 162^{\circ} = 18^{\circ}$180∘−162∘=18∘, then $n = 360^{\circ} \div 18^{\circ} = 20$n=360∘÷18∘=20 sides.

Grades 7–9 response: Does the above and, for "show that", checks back: sum $= (20-2)\times 180^{\circ} = 3240^{\circ}$=(20−2)×180∘=3240∘, and $3240^{\circ} \div 20 = 162^{\circ}$3240∘÷20=162∘ as required, confirming $n = 20$n=20 exactly.

Examiner-style commentary: routing through the exterior angle is the efficient method; the "show that" instruction rewards a clear verifying line.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: "In $\triangle ABC$△ABC, $AB = AC$AB=AC and $\angle BAC = 36^{\circ}$∠BAC=36∘. Work out $\angle ABC$∠ABC. Give a reason for each step."

Grades 3–4 response: Knows angles total $180^{\circ}$180∘ and may compute $180^{\circ} - 36^{\circ} = 144^{\circ}$180∘−36∘=144∘ but forgets to halve, or halves without explaining why.

Grades 5–6 response: States the triangle is isosceles so base angles are equal, then $\angle ABC = (180^{\circ} - 36^{\circ}) \div 2 = 72^{\circ}$∠ABC=(180∘−36∘)÷2=72∘.

Grades 7–9 response: Names the isosceles property explicitly ($AB = AC$AB=AC so $\angle ABC = \angle ACB$∠ABC=∠ACB), shows $2 \times \angle ABC + 36^{\circ} = 180^{\circ}$2×∠ABC+36∘=180∘, and concludes $\angle ABC = 72^{\circ}$∠ABC=72∘ with each reason written.

Examiner-style commentary: the most common lost mark is forgetting that the equal sides force equal base angles.

Common Mistakes

• Confusing "angles on a straight line $= 180^{\circ}$=180∘" with "angles at a point $= 360^{\circ}$=360∘".
• Treating corresponding or alternate angles as if they sum to $180^{\circ}$180∘; only co-interior angles do.
• Forgetting that $(n - 2) \times 180^{\circ}$(n−2)×180∘ works for irregular polygons too, not just regular ones.
• Setting one interior angle equal to the whole interior-angle sum when finding the number of sides.
• Giving a numerical answer with no named reason when the question says "give a reason".

Going Further

Once you are fluent, look for problems that combine facts: a regular polygon sitting on a parallel line, or an isosceles triangle formed by two radii inside a polygon. The exterior-angle route ($\dfrac{360^{\circ}}{n}$n360∘​) is almost always quicker than the interior-angle sum, so reach for it first when a regular polygon appears. These habits set you up directly for the circle theorems at Higher tier.

Exam Tips

• AO1 (use and apply): know the core facts cold — line $180^{\circ}$180∘, point $360^{\circ}$360∘, triangle $180^{\circ}$180∘, quadrilateral $360^{\circ}$360∘, exterior angles $360^{\circ}$360∘.
• AO2 (reason and communicate): "Give a reason for your answer" means name the fact (e.g. "alternate angles are equal"); a bare number scores the answer mark only.
• AO3 (solve problems): in multi-step questions, write each stage with its reason so the chain is visible; OCR command words here are "Work out", "Find" and "Show that".
• Look for arrows on a diagram — they tell you which lines are parallel and which angle rules apply.

Summary

• Angles on a line sum to $180^{\circ}$180∘; at a point to $360^{\circ}$360∘; vertically opposite angles are equal.
• Parallel-line angles: alternate (equal), corresponding (equal), co-interior (sum to $180^{\circ}$180∘).
• Triangle angles sum to $180^{\circ}$180∘; quadrilateral angles to $360^{\circ}$360∘; an exterior angle of a triangle equals the two opposite interior angles.
• Polygon interior-angle sum $= (n - 2) \times 180^{\circ}$=(n−2)×180∘; exterior angles sum to $360^{\circ}$360∘; for a regular polygon each exterior angle is $\dfrac{360^{\circ}}{n}$n360∘​.
• Always state the named geometric reason at every step.

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This content is aligned with the OCR GCSE Mathematics (J560) specification.