The concepts in this unit rely heavily on knowledge acquired previously, particularly for angles and scale drawings, so in this first section we revise these two topics.
In the diagram opposite, determine the size of each of the unknown angles.
Since
| c + 180° | = 180° (BCD is a straight line) |
| c | = 180° – 100° |
| c | = 80° |
| a | = 180° – (80° + 80°) |
| so a | = 20° |
In the diagram opposite, given that a = 65°, determine the size of each of the unknown angles.
| b = 180° – a | (angles on a straight line are supplementary, i.e. they add up to 180°) |
| b = 180° – 65° | |
| b = 115° | |
| c = a = 65° | (vertically opposite angles) |
| d = b = 115° | (corresponding angles, as the lines are parallel) |
| e = a = 65° | (corresponding angles) |
| f = a = 65° | (alternate angles) |
Draw an accurate plan of the car park which is sketched here. Use the scale
1 cm ≡ 10 m.
Estimate the distance AB.
The equivalent lengths are:
100 m ≡ 10 cm, 80 m ≡ 8 cm, 60 m ≡ 6 cm,
giving the following scale drawing:

In the scale drawing, AB = 11.7 cm, which gives an actual distance AB = 117 m in the car park.