Mathematics • Year 7 • Unit 3 • Lesson 4

Angle Sum of Triangles — Mixed Challenge

Combine the 180° rule with isosceles base angles, classification (by both angles and sides), and algebraic angle expressions. Spot a common substitution mistake and tackle an open-ended algebra puzzle.

Master · Mixed Challenge

1. Mixed problems — choose the right rule

Mix straight subtraction, isosceles, algebra and classification. Always state (angle sum of △). 2 marks each

1.1 Two angles of a triangle are 47° and 88°. Find the third.

1.2 An isosceles triangle has base angles of 72° each. Find the apex angle and classify by angles.

1.3 The angles of a triangle are x°, (x + 20)°, (x + 40)°. Find x and list all three angles.

1.4 The angles of a right-angled triangle are 90°, x° and (2x − 15)°. Find x.

1.5 An isosceles triangle has an apex of (2x)° and base angles of x° each. Find x.

1.6 Could a triangle have angles 60°, 60° and 60°? If yes, what type is it (combined angle and side classification)?

Stuck on 1.6? 60 + 60 + 60 = 180 ✓. All equal angles → all equal sides.

2. Find the mistake

Another Year 7 student tried to solve an isosceles triangle problem. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why, then re-do the working correctly. 3 marks

Student's problem: An isosceles triangle has an apex of 50°. Find each base angle.

Line 1: Let each base angle be B. Then 50 + B + B = 180 (angle sum of △).

Line 2: 50 + 2B = 180.

Line 3: 2B = 180 + 50 = 230.

Line 4: B = 230 ÷ 2 = 115°.

(a) Which line contains the mistake?

(b) Explain in one or two sentences why that line is wrong.

(c) Write out the corrected working in full, including the correct final value of B.

Stuck? To get 2B alone on one side, you must subtract 50 from BOTH sides — not add it.

3. Open-ended challenge — design triangles to spec

This question has many correct answers. 4 marks

3.1 Find three different sets of positive whole-number angles (none equal to 0°) that:

  • (i) sum to 180° (so they form a valid triangle);
  • (ii) include at least one angle that is a multiple of 15°;
  • (iii) make a triangle that is exactly ONE of: acute scalene, obtuse isosceles, or right isosceles — and you must produce at least one of each of those three types.

For each set: write the three angles, verify the sum is 180°, and label the type.

Bonus: Now write an algebraic angle expression of the form (x)°, (kx)° and ?° for the acute scalene one, where k is a whole number greater than 1, and the third angle is a constant (no x). Solve for x and verify the triangle.

Stuck on bonus? Try x + 2x + 60 = 180 → 3x = 120 → x = 40.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — 47°, 88°, ?

180 − 47 − 88 = 45° (angle sum of △).

1.2 — Base 72° each, find apex

Apex = 180 − 72 − 72 = 36°. All angles < 90° → acute-angled isosceles.

1.3 — x, x+20, x+40

x + (x+20) + (x+40) = 180 → 3x + 60 = 180 → 3x = 120 → x = 40. Angles: 40°, 60°, 80°. Check 40+60+80 = 180 ✓.

1.4 — Right triangle: 90°, x°, (2x−15)°

90 + x + (2x − 15) = 180 → 3x + 75 = 180 → 3x = 105 → x = 35. (Other angle: 2(35) − 15 = 55°. Check 90 + 35 + 55 = 180 ✓.)

1.5 — Isosceles, apex (2x)°, base x° each

2x + x + x = 180 → 4x = 180 → x = 45. (Angles: apex 90°, base 45° each — this is a right isosceles triangle.)

1.6 — 60°, 60°, 60°

Yes — sum is 180° ✓. All angles equal AND all < 90° → acute equilateral.

2 — Find the mistake

(a) The mistake is on Line 3.
(b) To isolate 2B in the equation 50 + 2B = 180, you must subtract 50 from both sides (giving 2B = 130), not add 50. Adding 50 makes the right side bigger instead of cancelling the 50 on the left.
(c) Corrected working:
Line 1: 50 + B + B = 180 (angle sum of △).
Line 2: 50 + 2B = 180.
Line 3 (fixed): 2B = 180 − 50 = 130.
Line 4 (fixed): B = 130 ÷ 2 = 65°. (Check: 50 + 65 + 65 = 180 ✓.)

3 — Open-ended challenge (sample answers)

Many valid answers; here is one valid set per type:
Acute scalene: 45°, 60°, 75° (sum 180 ✓; all < 90; all different; 45° and 75° are multiples of 15).
Obtuse isosceles: 30°, 30°, 120° (sum 180 ✓; one > 90; two equal; 30°/120° are multiples of 15).
Right isosceles: 45°, 45°, 90° (sum 180 ✓; one 90°; two equal; 45° and 90° are multiples of 15).
Bonus: x + 2x + 60 = 180 → 3x = 120 → x = 40. Angles: 40°, 80°, 60° — all < 90 and all different → acute scalene ✓. Check 40 + 80 + 60 = 180 ✓.

Marking: 1 mark for each of the three correctly-classified-and-verified sets; 1 mark for the bonus algebraic working.