Angles in Geometric Figures

Finding unknown angles in composite figures involving triangles and all special quadrilaterals — parallelogram, rhombus, trapezium.

triangle angle sumparallelogram/rhombustrapezium P1 · no calc P2 · calc

Triangle: a + b + c = 180° Straight line: angles sum to 180° Angles at a point: sum to 360° Vertically opposite angles are equal

Parallelogram

Opposite angles equal
Adjacent angles add to 180°
Opposite sides equal & parallel

Rhombus

All sides equal
Opposite angles equal
Adjacent angles add to 180°

Trapezium

One pair of parallel sides
Co-interior angles add to 180°
(between the parallel sides)

Isosceles triangle

Two equal sides
Base angles are equal
Angle sum = 180°

Part A — Triangles

Question 1 — Short answer (2 marks) Paper 1 · no calculator Isosceles triangle

In the figure, ABC is an isosceles triangle where AB = AC. Angle BAC = 48°. Find angle ABC.

AB = AC→ angle ABC = angle ACB (base angles of isosceles triangle)

Sum of anglesangle ABC + angle ACB + 48° = 180°

2 × angle ABC= 180° − 48° = 132°

Angle ABC= 132° ÷ 2 = 66°

Angle ABC = 66°

Always state the property used: "base angles of isosceles triangle are equal." Examiners award marks for reasoning, not just the final number.

Question 2 — Short answer (2 marks) Paper 1 · no calculator Triangle on straight line

In the figure, BCD is a straight line. Angle ABC = 115° and angle ACB = 38°. Find angle ACD.

Angle BAC= 180° − 115° − 38° = 27° (angle sum of triangle)

Angle ACD= 180° − 38° = 142° (angles on a straight line)

Angle ACD = 142°

Exterior angle theorem shortcut: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles. Here, angle ACD is exterior to triangle ABC at vertex C. The two non-adjacent interior angles are angle ABC (115°) and angle BAC (27°). So angle ACD = 115° + 27° = 142° — same answer, one step. The straight-line method (180° − 38°) is more direct here, but both approaches are valid and give the same result.

Part B — Parallelogram & Rhombus

Question 3 — Short answer (2 marks) Paper 1 · no calculator Parallelogram

ABCD is a parallelogram. Angle DAB = 72°. Find angle ABC and angle BCD.

Angle ABC= 180° − 72° = 108° (adjacent angles in parallelogram sum to 180°)

Angle BCD= angle DAB = 72° (opposite angles in parallelogram are equal)

Angle ABC = 108° | Angle BCD = 72°

Two key parallelogram properties: (1) opposite angles are equal, (2) adjacent angles are supplementary (add to 180°). A rhombus has the same angle properties — only difference is all four sides are equal.

Question 4 — Structured (3 marks) Paper 1 · no calculator Rhombus + triangle composite

In the figure, ABCD is a rhombus and ABE is a triangle with EB = EA. Angle DAB = 126° and angle ABE = 33°. Find angle AEB.

Angle ABC= 180° − 126° = 54° (adjacent angles in rhombus sum to 180°)

EB = EA→ triangle ABE is isosceles, so angle EAB = angle ABE = 33°

Angle AEB= 180° − 33° − 33° = 114°

Angle AEB = 114°

Key step: EB = EA means triangle ABE is isosceles with the two equal angles at B and A. So angle EAB = angle ABE = 33°. The rhombus angle was a distractor here — angle DAB is not needed to find angle AEB once you identify the isosceles triangle.

Part C — Trapezium (hardest type)

Question 5 — Long answer (4 marks) Paper 2 · calculator Trapezium + triangle composite Distinction level

In the figure, ABCD is a trapezium with AD parallel to BC. Angle DAB = 58°, angle BCD = 110°, and E is a point on AD with angle CBE = 42°. Find angle AEB.

Angle ABC= 180° − 58° = 122° (co-interior angles, AD ∥ BC, sum to 180°)

Angle ABE= angle ABC − angle CBE = 122° − 42° = 80°

Angle BAE= angle DAB = 58° (E lies on AD, so angle BAE is the same as angle DAB)

Angle AEB= 180° − 80° − 58° = 42°

Angle AEB = 42°

The key property for trapeziums: Co-interior angles (also called allied angles) between parallel lines sum to 180°. Here AD ∥ BC, so angle DAB + angle ABC = 180°. This is the one trapezium property that must be memorised — it is the only way to link the two parallel sides.