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

Finding Unknown Angles in Composite Figures

PSLE Geometry practice · Primary 6

Each question below combines two or more shapes, and finding the unknown angle takes a chain of angle properties rather than a single fact — the multi-step reasoning the PSLE rewards. Every solution can be reached without drawing any extra lines. Tap Show working to check your method, not just your answer — in the exam, marks are given for the correct reason at each step.

Triangle: angles sum to 180° Straight line: 180° Angles at a point: 360° Vertically opposite angles equal

Equilateral triangle

All 3 sides equal
All 3 angles = 60°

Isosceles triangle

Two equal sides
Base angles (opposite the equal sides) are equal

Rectangle & square

All 4 angles = 90°
Opposite sides equal & parallel
(diagonal properties not tested)

Parallelogram

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

Rhombus

All 4 sides equal
Opposite angles equal
Adjacent angles add to 180°

Trapezium

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

All diagrams are not drawn to scale. Tick marks (▏) show sides that are equal in length.

Paper 1 — Multiple-choice & short-answer · no calculator

Question 1 Paper 1 · Booklet A · MCQ · 2 marks Equilateral + isosceles 5 steps

In the figure, BCD is an equilateral triangle. ABD is an isosceles triangle with AD = BD and ∠ABD = 52°. Find ∠DCA.

(1) 22°

(2) 26°

(3) 38°

(4) 52°

∠DAB= ∠ABD = 52° (AD = BD, so triangle ABD is isosceles → base angles equal)

∠ADB= 180° − 52° − 52° = 76° (angle sum of triangle ABD)

AD = CDBD = CD (all sides of equilateral triangle BCD are equal); since AD = BD, AD = CD → triangle ACD is isosceles

∠ADC= ∠ADB + ∠BDC = 76° + 60° = 136° (∠BDC = 60° from equilateral triangle BCD)

∠DCA= (180° − 136°) ÷ 2 = 22° (base angles of isosceles triangle ACD; angle sum)

∠DCA = 22° → option (1)

The hidden step: nobody tells you triangle ACD is isosceles. You have to spot that AD = BD = CD — the given AD = BD links to BD = CD from the equilateral triangle. Miss that equal length and the question looks unsolvable. Stopping at ∠ADB = 76° gives the distractor 52° (option 4).

Question 2 Paper 1 · Booklet A · MCQ · 2 marks Two isosceles triangles 4 steps

ABC is an isosceles triangle with AB = AC. D lies on AC such that BD = BC. Given ∠ABD = 30°, find ∠BAC.

(1) 30°

(2) 36°

(3) 40°

(4) 50°

Equal angles∠ABC = ∠ACB (AB = AC). D is on AC, so ∠BCD = ∠ACB. Therefore ∠BCD = ∠ABC.

∠BDC= ∠BCD (BD = BC, so triangle BDC is isosceles). So ∠BDC = ∠ABC as well.

Triangle BDCLet ∠ABC be 1 part. Then ∠DBC = (1 part − 30°), ∠BDC = ∠BCD = 1 part. Angle sum: (1 part − 30°) + 1 part + 1 part = 180° → 3 parts = 210° → 1 part = 70°

∠BAC= 180° − 70° − 70° = 40° (∠ABC = 70°; angle sum of triangle ABC)

∠BAC = 40° → option (3)

The breakthrough: three different angles — ∠ABC, ∠BCD and ∠BDC — all turn out to be equal. That lets the triangle BDC angle sum become "3 parts − 30° = 180°". Without spotting that, there are too many unknowns to make progress.

Question 3 Paper 1 · Booklet B · Short-answer · 2 marks Parallelogram + diagonal 3 steps

ABCD is a parallelogram. The diagonal BD divides ∠ABC so that ∠ABD is twice ∠DBC. Given ∠BCD = 66°, find:

(a) ∠BDC (b) ∠DBC

(a) ∠BDC

∠ABC= 180° − 66° = 114° (∠ABC and ∠BCD are adjacent angles of the parallelogram → add up to 180°)

Split 2 : 1∠ABD + ∠DBC = 114°, with ∠ABD = 2 × ∠DBC → 3 parts = 114° → ∠ABD = 76°

∠BDC= ∠ABD = 76° (AB ∥ DC, so ∠ABD and ∠BDC are alternate angles)

∠BDC = 76°

(b) ∠DBC

∠DBC= 114° ÷ 3 = 38° (1 part)

∠DBC = 38°

Trap: ∠ABC comes from the adjacent-angle rule (180° − 66°), not the opposite-angle rule. After splitting it into 3 equal parts, part (a) is fastest by alternate angles — but the triangle BDC route (180° − 66° − 38°) gives the same 76°.

Paper 2 — Structured & long-answer · calculator allowed · show working

Question 4 Paper 2 · Structured · 4 marks Rectangle + intersecting lines 3 steps

ABCD is a rectangle. E lies on AD. The diagonal AC meets BE at F. Given ∠ABE = 50° and ∠ACB = 28°, find ∠BFC

∠BAC= 90° − 28° = 62° (∠ABC = 90° in the rectangle; angle sum of triangle ABC with ∠ACB = 28°)

∠ABF= ∠ABE = 50° (F lies on BE)

∠AFB= 180° − 62° − 50° = 68° (angle sum of triangle ABF)

∠BFC= 180° − 68° = 112° (A, F, C lie on the straight line AC → angles on a straight line)

∠BFC = 112°

Trap: the only "rectangle" fact you need is the 90° corner at B (to get ∠BAC = 62°). You never need rectangle diagonal properties — they aren't part of the syllabus. Then ∠AFB is a plain triangle, and ∠BFC follows from the straight line AC.

Question 5 Paper 2 · Structured · 4 marks Rhombus + isosceles + straight line 4 steps

In the figure, ABCD is a rhombus and DCE is an isosceles triangle with DC = DE. B, C and E lie on a straight line. ∠ADC = 76°.

(a) Find ∠DCE. (b) Find ∠CDE.

(a) ∠DCE

∠BCD= 180° − 76° = 104° (∠ADC and ∠BCD are adjacent angles of the rhombus → add up to 180°)

∠DCE= 180° − 104° = 76° (B, C, E lie on a straight line → angles on a straight line add up to 180°)

∠DCE = 76°

(b) ∠CDE

∠DEC= ∠DCE = 76° (DC = DE, so triangle DCE is isosceles → base angles ∠DCE and ∠DEC are equal)

∠CDE= 180° − 76° − 76° = 28° (angle sum of triangle DCE)

∠CDE = 28°

Trap: the equal sides are DC and DE, so the two equal angles are at C and E (∠DCE and ∠DEC) — not at D. ∠CDE is the apex angle, which is whatever is left over.

Question 6 Paper 2 · Structured · 4 marks Trapezium + isosceles 4 steps

In the figure, ABCD is a trapezium with AB parallel to DC. BC = BD. ∠BCD = 54° and ∠DAB = 100°.

(a) Find ∠DBC. (b) Find ∠ABD.

(a) ∠DBC

∠BDC= ∠BCD = 54° (BC = BD, so triangle BCD is isosceles → base angles ∠BDC and ∠BCD are equal)

∠DBC= 180° − 54° − 54° = 72° (angle sum of triangle BCD)

∠DBC = 72°

(b) ∠ABD

∠ABC= 180° − 54° = 126° (AB ∥ DC, so ∠ABC and ∠BCD are co-interior angles → add up to 180°)

∠ABD= ∠ABC − ∠DBC = 126° − 72° = 54°

∠ABD = 54°

Trap: ∠ABC must come from the parallel sides (co-interior with the 54°). It is tempting — but wrong — to assume the trapezium is "isosceles". Only after finding ∠ABC = 126° can you subtract the 72° from part (a).

Question 7 Paper 2 · Long-answer · 4 marks Rhombus + equilateral 4 steps

In the figure, ABCD is a rhombus and BCE is an equilateral triangle, with E lying inside angle BCD (between CB and CD). ∠DAB = 140°. Find ∠CDE.

∠BCD= ∠DAB = 140° (opposite angles of a rhombus are equal)

∠BCE= 60° (triangle BCE is equilateral)

∠DCE= ∠BCD − ∠BCE = 140° − 60° = 80° (CE lies inside angle BCD)

CE = CDCB = CD (all sides of a rhombus are equal) and CE = CB (equilateral) → CE = CD, so triangle CDE is isosceles

∠CDE= (180° − 80°) ÷ 2 = 50° (base angles of isosceles triangle CDE; angle sum)

∠CDE = 50°

The key link: CE = CB = CD. The equilateral triangle gives CE = CB; the rhombus gives CB = CD. Only by joining these does triangle CDE become isosceles — which is what lets you split (180° − 80°) into two equal base angles.