Module 3 · Assignment 3

Assignment — questions & solutions

Original Wolsey Hall assignment questions (with diagrams redrawn) followed by worked answers. Allow ±2° tolerance on protractor / construction measurements.

Question 1 — Find a, b, c and d (8 marks)

The diagram is not drawn to scale. Show clearly how you found each angle, using correct mathematical terms (e.g. alternate angles).

Question 1 — find angles a, b, c, d.

a = 30°

150° and the angle next to it on the top line are angles on a straight line: 180° − 150° = 30°. That 30° interior angle of the triangle and angle a are alternate angles (between the parallel lines, on opposite sides of the left transversal), so a = 30°.

b = 50°

b is vertically opposite the marked 50°, so b = 50°.

c = 50°

b and c are alternate angles across the right transversal (between the parallel lines, opposite sides), so c = b = 50°.

d = 100°

a, d and c sit on the bottom line above it — angles on a straight line sum to 180°.
30° + d + 50° = 180° → d = 100°.
Or: angles in the triangle 30° + 50° + d = 180°.

Question 2 — Measure these angles (2 marks)

Question 2 — measure angles (i) and (ii) with a protractor.

≈ 150°

Obtuse — measured from the horizontal arm round to the down-right arm. Accept 148–152°.

≈ 45°

Acute — the open mouth of the ">" shape. Accept 43–47°.

Question 3 — Regular polygon (4 marks)

A regular polygon has an interior angle of 172°. Find the exterior angle and work out how many sides it has, making your method clear.

Exterior angle = 8°. Number of sides = 45.

Interior + exterior = 180° (angles on a straight line at each vertex).
Exterior = 180° − 172° = 8°.
For a regular polygon, exterior × n = 360°.
n = 360 ÷ 8 = 45 sides.

Question 4 — Quadrilateral (5 marks)

(a) Show how you could calculate the angle marked f. (b) Measure each side, write the lengths on the diagram and construct the quadrilateral afresh.

Question 4 — find f, then construct an accurate copy.

f = 130°

Angles in a quadrilateral sum to 360°.
f = 360° − 100° − 68° − 62° = 130°.

Construction — see method.

1. Measure each side of the printed quadrilateral with a ruler and write the lengths on the diagram.
2. On fresh paper, accurately draw the bottom side.
3. Use a protractor at each end to draw rays at the given angles (68° at the bottom-left, 62° at the bottom-right).
4. Measure along each ray to mark the next vertex at the correct distance.
5. Join the two upper vertices to complete the shape.
Check: top-left should measure 100°, top-right 130°.

Question 5 — Map scale (3 marks)

A map has a scale of 1 : 250 000. How many centimetres on the map is a journey of 40 km?

16 cm on the map

Convert real distance to cm:
40 km = 40 × 1000 × 100 = 4 000 000 cm.
Map distance = real ÷ scale factor = 4 000 000 ÷ 250 000 = 16 cm.

Question 6 — Three towns (6 marks)

Diagram: Alton (A) at the bottom; Bolton (B) is to the upper-right with AB = 6 cm; Carlton (C) is to the upper-left. Each town has a North line. Scale: 1 cm to 2.5 km.

Question 6 — Alton, Bolton, Carlton.

Alton to Bolton = 15 km

AB on the diagram = 6 cm. Real distance = 6 × 2.5 = 15 km.

6bi

Mark the angle at B, measured clockwise from B's North line round to BA (the line going down-left back to Alton).

6bii

Bearing of Alton from Bolton = 226°

Forward bearing of B from A = 046°.
Back bearing = forward + 180° (since forward < 180°).
= 046° + 180° = 226°.

6ci

Mark the angle at C, measured clockwise from C's North line all the way round to CA (the line going down-right back to Alton).

6cii

Bearing of Alton from Carlton = 118°

Forward bearing of C from A: C is 62° anticlockwise of N at A, so the bearing of C from A = 360° − 62° = 298°.
Back bearing = forward − 180° (since forward > 180°).
= 298° − 180° = 118°.

Question 7 — Boat journey (4 marks)

A boat starts at A. It travels 7 km on a bearing of 135° to B. From B it travels 12 km on a bearing of 250° to C. Using a scale of 1 cm : 1 km, draw a diagram to show these bearings and journeys.

Construction — method below.

1. Mark point A. Draw a faint North line straight up from A.
2. Use a protractor: measure 135° clockwise from A's North line and draw a ray.
3. Along that ray, measure 7 cm and mark B.
4. At B, draw a new (parallel) North line straight up.
5. Measure 250° clockwise from B's North line — i.e. into the south-west region — and draw a ray.
6. Along that ray, measure 12 cm and mark C.
Check: 135° points roughly south-east; 250° points roughly south-west.

Question 8 — Triangle PQR (4 marks)

Construct the triangle below using a ruler and protractor. PQ = 65 m, angle P = 42°, angle Q = 69°. Use a scale of 1 cm : 10 m. (Diagram is not to scale.)

Question 8 — triangle PQR (ASA).

Construction.

1. Draw PQ = 6.5 cm.
2. At P, draw a ray at 42° above PQ.
3. At Q, draw a ray at 69° above PQ.
4. Where the two rays meet is R.

Angle R = 69°

Angles in a triangle sum to 180°: R = 180° − 42° − 69° = 69°. Measured value should be within ±2°.

Isosceles triangle

Two angles equal (Q = R = 69°), so the two sides opposite them are equal (PR = PQ). A triangle with two equal angles is isosceles.

Question 9 — Triangle SUT (4 marks)

Construct the triangle below using a ruler and protractor. SU = 4 cm, ST = 6 cm, angle S = 45°. Scale 1 cm : 10 m. (Diagram is not to scale.)

Question 9 — triangle SUT (SAS).

Construction.

1. Draw the base ST = 6 cm.
2. At S, draw a ray at 45° above ST.
3. Measure 4 cm along that ray and mark U.
4. Join U to T to finish.

Angle U ≈ 93°

Measured with a protractor on the construction. Accept 91–95°.

Use the angle-sum-of-a-triangle rule.

Angles in a triangle sum to 180°.
T = 180° − S − U = 180° − 45° − 93° ≈ 42°.

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