Mathematics • Year 7 • Unit 3 • Lesson 1
Angles — Mixed Challenge
Mix every angle relationship from Lesson 1: classify by size, name angles using vertex notation, use complementary, supplementary, vertically opposite and angles-at-a-point rules. Spot a common error, then tackle an open-ended puzzle.
1. Mixed problems — choose the right rule
Each question requires you to choose which relationship applies BEFORE you calculate. Show working and a reason. 2 marks each
1.1 Classify each angle as acute, right, obtuse, straight or reflex: (a) 15°, (b) 90°, (c) 180°, (d) 215°.
1.2 Find the complement of 41° and the supplement of 41°. (Two answers.)
1.3 Two straight lines cross. One of the four angles is 153°. State the other three with a reason for each.
1.4 Three angles sit side-by-side on a straight line at one point. They are 40°, 2x° and 60°. Find x.
1.5 Five angles meet at a single point. Four are 90°, 60°, 75° and 90°. Find the fifth.
1.6 In ∠ABC, what is the vertex? Then explain why "longer rays" don't change the size of the angle.
2. Find the mistake
Another Year 7 student has tried to find an unknown angle. Their working is shown below — exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's problem: Two straight lines cross. One angle is 105°. Find the angle adjacent to it along one of the lines.
Line 1: The angle adjacent on the same line is vertically opposite to 105°.
Line 2: Vertically opposite angles are equal.
Line 3: Therefore the adjacent angle = 105°.
(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 answer and the geometric reason.
Stuck? "Adjacent on the same line" means the two angles together form a straight line — so they are supplementary, not vertically opposite.3. Open-ended challenge — design a six-way junction
This question has many correct answers. Show your work. 4 marks
3.1 An urban designer is planning a six-way pedestrian junction (six paths meeting at one centre point). Choose six positive whole-number angles that:
(i) sum to 360° (so all six paths fit around the point);
(ii) include at least one angle that is acute, at least one that is obtuse, and exactly one that is a right angle;
(iii) include at least two angles that are vertically opposite across the junction (so two pairs of opposite angles are equal).
List your six angles, label which pair is vertically opposite, and verify the sum is 360°.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Classify
(a) 15° → acute. (b) 90° → right. (c) 180° → straight. (d) 215° → reflex (between 180° and 360°).
1.2 — Complement and supplement of 41°
Complement = 90 − 41 = 49°. Supplement = 180 − 41 = 139°.
1.3 — Two lines cross, one angle = 153°
Vertically opposite = 153° (vert. opp. ∠s). Adjacent on line = 180 − 153 = 27° (∠s on str. line). Fourth = 27° (vert. opp. ∠s).
1.4 — 40° + 2x° + 60° on a straight line
∠s on str. line sum to 180°. 40 + 2x + 60 = 180 → 2x = 80 → x = 40. (The middle angle is 80°.)
1.5 — Five angles at a point
90 + 60 + 75 + 90 = 315. Fifth = 360 − 315 = 45° (∠s at a pt.).
1.6 — Vertex of ∠ABC
Vertex = B (middle letter rule). Angle size depends only on the amount of rotation between the two rays — not the length of the rays. Two short rays at 30° apart and two long rays at 30° apart are the same angle.
2 — Find the mistake
(a) The mistake is on Line 1.
(b) The angle adjacent to 105° along the same straight line sits next to it and forms a straight line — so the two angles are supplementary (sum 180°), not vertically opposite. Vertically opposite is the angle across the X, not the one beside it.
(c) Corrected working:
Line 1: The adjacent angle and the 105° angle together form a straight line.
Line 2: ∠s on str. line sum to 180°.
Line 3: Adjacent = 180 − 105 = 75°.
3 — Open-ended six-way junction (sample solution)
Many valid answers. One example:
Angles around the centre, going clockwise: 90°, 100°, 30°, 90°, 50°, ? — sum so far = 360, so the last one is 0° (not allowed). Try again:
Sample valid set: 90°, 100°, 30°, 100°, 20°, 20°. Sum = 90 + 100 + 30 + 100 + 20 + 20 = 360° ✓.
Has one right angle (90°), acute angles (30°, 20°, 20°), an obtuse angle (100°), and the two 100°s sit directly opposite each other (vertically opposite). The two 20°s are also vertically opposite. ✓
Another sample: 110°, 110°, 90°, 25°, 25°, ? = 360 needs 360 − 360 = 0 (no). Try 110°, 110°, 90°, 20°, 15°, 15°. Sum = 360°. ✓ Two 110°s opposite each other, two 15°s opposite each other.
Marking: 1 for sum = 360°; 1 for at least one acute, one obtuse, one right angle; 1 for clearly labelled vertically-opposite pair; 1 for clear verification of the sum.