Module 3 · Answers

Answers & working

Workings shown so you can see where each answer comes from.

Section A — Easy

01
70°
360 − 90 − 90 − 110 = 70.
02
(a) 65°    (b) 65°    (c) 115°
Corresponding and alternate equal; co-interior add to 180°.
03
540°
(5 − 2) × 180 = 540.
04
9 sides
360 ÷ 40 = 9.
05
(a) 090°    (b) 225°    (c) 000° (or 360°)
06
16m
8 × 200 = 1600cm = 16m.

Section B — Medium

07
x = 60
x + 2x + (x + 30) + 90 = 360
4x + 120 = 360
4x = 240
x = 60.
08
135°
Sum = (8 − 2) × 180 = 1080°. Each = 1080 ÷ 8 = 135°.
09
x = 32
Co-interior angles: 3x + (2x + 20) = 180
5x = 160
x = 32.
10
230°
Back-bearing = original + 180° (when original < 180°). 050 + 180 = 230.
11
1.1 km
4.4 × 25 000 = 110 000cm = 1100m = 1.1 km.

Section C — Hard

12
125° each
Sum (8 sides) = (8 − 2) × 180 = 1080°.
Known: 130 + 140 + 150 + 160 = 580°.
Remaining: 1080 − 580 = 500°, shared between 8 − 4 = 4 equal angles.
500 ÷ 4 = 125°.
13
10 sides (decagon)
Exterior angle = 180 − 144 = 36°.
n = 360 ÷ 36 = 10.
14
Approx 270° (due west)
First leg: 8km on 060°. Second leg: 6km on 150° (turning 90° clockwise). Result is east + south-east of start. By scale drawing, the start sits roughly due west of the final position, so the bearing of the start from the final point is about 270°. Exact value via Pythagoras / trig is approximately 273°.
15
Largest angle ≈ 82.8° (opposite the 6cm side)
Construction:
1. Draw the 6cm side (the longest) as a baseline.
2. From one end, draw an arc of radius 5cm.
3. From the other end, draw an arc of radius 4cm.
4. Where the arcs intersect is the third vertex.
5. Join up. Measure the angle at the apex (opposite the 6cm side) — about 82°.
The largest angle is always opposite the longest side.
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