📏 Maths · Paper 2 · Geometry (Angles)

The folded triangle,
angle by angle

PSLE 2017 Paper 2, Question 17 is a favourite “hardest question” for a reason: it chains four separate ideas — isosceles base angles, the angle sum of a triangle, angles on a straight line, and what folding actually does to an angle. Get comfortable with the chain here, and every fold-and-angle question after this gets easier.

PSLE 2017 · Paper 2 · Q17 5 marks (2 + 3) Isosceles triangles + folding
The question

📝 PSLE 2017, Paper 2, Question 17

Minah has a triangular piece of paper ABC with BA = BC, ∠ABC = 84° and ∠CDE = 67°. ADC and BEC are straight lines. She folds the paper along the line DE, as shown below.

(a) Find ∠x.    (b) Find ∠y.

PSLE 2017 Paper 2 Question 17, the figure from the paper Left, before folding: triangle ABC with BA equal to BC and angle ABC 84 degrees. D lies on AC and E lies on BC, joined by a dashed crease DE, with angle CDE 67 degrees and a curved arrow showing the flap swinging to the left. An arrow points right to the second panel, drawn at the same scale, showing the paper after folding along DE, with the 84 degree angle at B labelled outside the figure and joined to it by a curved arrow, and the unknown angles x and y marked. 84° 67° A B C D E 84° x y Before folding After folding

💡 How to start any folding question

Folding is a reflection in the crease DE. Paper doesn’t stretch, so every angle inside the flap is copied exactly — it just lands somewhere new.

That means the first move is never a calculation. It is one of these two questions:

  • Is the unknown a folded copy of something? If so, go back to the before picture and find the original instead. (This solves ∠x.)
  • If not, which triangle does it sit in, and what is the one missing angle there? Then go and get that angle. (This solves ∠y.)
Worked solution

Part (a): Find ∠x

2 marks. Before touching any numbers, work out what ∠x actually is.

🔍 Start at the end: what is ∠x?

∠x sits at E, between the crease ED and the folded edge EC′. Before the fold, that same angle was ∠DEC — the angle at E inside triangle CDE. Folding just carried it across the crease unchanged. So the question “find ∠x” is really the question “find ∠DEC”, and that is an ordinary triangle problem. Everything below is just getting there.

  1. 1 Rewrite the question: ∠x = ∠DEC.

    Folding is a reflection in DE, and reflections don’t change the size of an angle. So instead of chasing ∠x in the messy folded picture, go back to the clean before picture and find ∠DEC there.

  2. 2 To use triangle CDE, first find ∠DCE.

    Triangle CDE already gives us ∠CDE = 67°. We need one more angle. Since BA = BC, triangle ABC is isosceles, so its base angles are equal: ∠BAC = ∠BCA = (180° − 84°) ÷ 2 = 48°. D lies on AC, so ∠DCE is that same angle: 48°.

  3. 3 Angle sum of triangle CDE.

    ∠DEC = 180° − 48° − 67° = 65°.

  4. 4 Carry it back across the fold.

    ∠x is that angle reflected, so ∠x = ∠DEC = 65°.

84° 67° 48° 48° 65° A B C D E

Answer — part (a)

∠x = 65°

Worked solution

Part (b): Find ∠y

3 marks — and again, the first move is working out which angle would unlock it.

🔍 Start at the end: what would unlock ∠y?

Look at where ∠y sits. The folded edge DC′ crosses edge AB — call that crossing point F (the exam paper doesn’t name it, so mark it in yourself). ∠y is ∠AFD, an angle inside triangle ADF.

In that triangle we already know ∠DAF = ∠BAC = 48°. So there is exactly one missing piece: ∠ADF — which is ∠C′DA, the angle the folded edge makes with DA. Find ∠C′DA and ∠y drops out of the angle sum. That is the whole plan.

  1. 1 Identify the triangle ∠y lives in.

    Mark F where DC′ crosses AB. Then ∠y = ∠AFD, inside triangle ADF, whose other two angles are ∠DAF and ∠ADF (= ∠C′DA).

  2. 2 One angle of that triangle is already known.

    ∠DAF is just the original base angle at A, untouched by the fold: 48°.

  3. 3 Unlock the missing angle, ∠C′DA.

    ADC is a straight line, so the angles at D add to 180°. Turning from DC round to DA you pass through two equal 67° angles: ∠CDE = 67° (given), and ∠EDC′ = 67° (its mirror image, because the fold copied it across the crease). What is left over is ∠C′DA:

    ∠C′DA = 180° − 67° − 67° = 46°

  4. 4 Angle sum of triangle ADF.

    ∠y = 180° − 48° − 46° = 86°.

84° x = 65° y = 86° 46° 48° A B D E F C′
The same after-folding figure, with the points the exam paper leaves unnamed — A, D, F and C′ — marked in.

Answer — part (b)

∠y = 86°

Why this question is hard

Four skills, chained in one question

This is what pushes a question to 5 marks and a top difficulty rating — not any single step, but needing all four without a prompt.

  • Isosceles triangle base angles.BA = BC tells you two angles are equal before you're told anything about D or E.
  • Angle sum of a triangle (twice).Once in triangle CDE for part (a), and again in the unlabelled triangle ADF for part (b).
  • Angles on a straight line.ADC being straight is what lets you strip off the two 67° angles at D to reach 46° in part (b).
  • Recognising what a fold preserves.Angles and lengths carry over unchanged — only their position on the page changes.

★ Teacher’s power tip

Notice that neither part of this question starts with a calculation. Both start by naming the unknown: ∠x is ∠DEC folded over; ∠y is the third angle of triangle ADF. Once the unknown has a name, the arithmetic is P5 work. Students who dive straight into computing angles may produce three correct numbers that lead nowhere.

A useful habit: whenever a folded edge crosses another edge without a labelled point there, mark that crossing yourself and give it a letter (like F here). You cannot name the triangle your unknown sits in until its corners have names.