Mathematics • Year 7 • Unit 3 • Lesson 6

Triangle Problem Solving

Build fluency with the four-step strategy: KNOW → WANT → WHICH TRIANGLE → RULE + REASON. Pull together angle sum (180°), exterior angle theorem (e = a + b), isosceles base angles, and equilateral angles, and write a reason in brackets after every line.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. The four-step strategy is shown in order with a reason in brackets at the end of each line of working.

Problem. An isosceles triangle has one base angle of 70°. Find the apex angle.

Step 1 — KNOW.

Triangle is isosceles. One base angle = 70°.

Reason: a base angle is one of the two equal angles opposite the equal sides.

Step 2 — WANT.

Apex angle (the angle between the two equal sides).

Step 3 — WHICH TRIANGLE + RULE.

Other base angle = 70° (base ∠s isos.)

Reason: equal sides → equal opposite (base) angles.

Step 4 — REASON + COMPUTE.

Apex = 180 − 70 − 70 (∠ sum of △)

Apex = 40°

Check: 70 + 70 + 40 = 180° ✓

Answer: Apex angle = 40°.

Stuck? Revisit lesson § "Isosceles Problems" — write down the four-step strategy and tick each step as you finish it.

2. We do — fill in the missing steps

Same structure as Section 1, but with the working faded. Fill in each blank. 4 marks

Problem. A triangle has angles 2x°, 3x° and 5x°. Find x and state all three angles.

Step 1 — KNOW. Three angles are written in terms of x.

Step 2 — WANT. The value of x, then each angle.

Step 3 — WHICH TRIANGLE + RULE. All three angles belong to one triangle, so use:

2x + 3x + 5x = _______ (∠ sum of △)

Step 4 — Collect like terms:

_______ x = 180

Step 5 — Solve:

x = 180 ÷ _______ = _______

Step 6 — Substitute back:

Angles = 2 × _______ ° , 3 × _______ ° , 5 × _______ ° = _______ ° , _______ ° , _______ °

Check: _______ + _______ + _______ = 180° ✓

Stuck? Revisit lesson § "Algebraic Angles" — collect coefficients of x BEFORE you divide.

3. You do — independent practice

Show your working in the space under each problem. The first four are foundation, the middle two are standard, and the last two are extension. Always write the reason in brackets.

Foundation — one rule at a time

3.1 Isosceles triangle has one base angle of 55°. Find the apex angle. 1 mark

3.2 Angles of a triangle are x°, 2x°, 3x°. Find x. 1 mark

3.3 Isosceles triangle has apex 40°. Find each base angle. 1 mark

3.4 State each angle of an equilateral triangle, and give the reason. 1 mark

Standard — two rules in one problem

3.5 Triangle has interior angles 40°, 70°, 70°. Find the exterior angle at one of the 70° vertices. 2 marks

3.6 Isosceles triangle has apex angle of 120°. Find each base angle, then find the exterior angle at the apex. 2 marks

Extension — chain the rules

3.7 Triangles ABD and BCD share side BD with A, B, C on a straight line. In triangle ABD: ∠DAB = 50°, ∠ABD = 80°. In triangle BCD: ∠BCD = 40°. Find ∠ADB, ∠DBC and ∠BDC. Give a reason on each line. 3 marks

3.8 A triangle has angles (2x + 10)°, (3x − 10)° and (x + 60)°. Find x, then state every angle. 2 marks

Stuck on 3.8? Add coefficients of x separately from the constants, then move the constant over and divide.

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Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (2x + 3x + 5x = 180)

RHS = 180. Coefficient: 10x = 180, so x = 180 ÷ 10 = 18.
Angles = 2 × 18°, 3 × 18°, 5 × 18° = 36°, 54°, 90°.
Check: 36 + 54 + 90 = 180° ✓

3.1 — Isosceles, base 55°

Other base = 55° (base ∠s isos.). Apex = 180 − 55 − 55 = 70° (∠ sum of △).

3.2 — x, 2x, 3x

x + 2x + 3x = 180 (∠ sum of △). 6x = 180, so x = 30.

3.3 — Apex 40°

Each base = (180 − 40) ÷ 2 = 140 ÷ 2 = 70° (base ∠s isos. + ∠ sum of △).

3.4 — Equilateral

All three sides equal, all three angles equal, and the angles sum to 180°. So each angle = 180 ÷ 3 = 60° (equilateral / ∠ sum of △).

3.5 — Exterior at 70° vertex

By the exterior angle theorem, the exterior angle = sum of the two remote interior angles = 40 + 70 = 110° (ext. ∠ of △).
Check: exterior + interior at that vertex = 110 + 70 = 180° ✓ (angles on a straight line).

3.6 — Isosceles apex 120°

Each base = (180 − 120) ÷ 2 = 30° (base ∠s isos. + ∠ sum of △).
Exterior angle at apex = 180 − 120 = 60° (∠s on str. line). Cross-check: 30 + 30 = 60° ✓ (ext. ∠ of △).

3.7 — Two triangles sharing BD

∠ADB = 180 − 50 − 80 = 50° (∠ sum of △ABD).
∠DBC = 180 − 80 = 100° (∠s on str. line at B).
∠BDC = 180 − 100 − 40 = 40° (∠ sum of △BCD).

3.8 — (2x + 10) + (3x − 10) + (x + 60) = 180

Collect: 6x + 60 = 180 (∠ sum of △). So 6x = 120, x = 20.
Angles = 2(20) + 10 = 50°, 3(20) − 10 = 50°, 20 + 60 = 80°.
Check: 50 + 50 + 80 = 180° ✓