Edexcel GCSE Maths (1MA1), Higher tier · Geometry and measures
Practise Bearings and scale drawings. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
The bearing of \(B\) from \(A\) is 070°. Tom says, 'The bearing of \(A\) from \(B\) is 110°.'
(a) Explain why Tom is wrong.[1]
(b) Work out the bearing of \(A\) from \(B\).[1]
Show the answer and mark scheme
(a)Answer: The bearing back is in the opposite direction, so you add 180°: \(70 + 180 = 250\). Tom worked out \(180 - 70\).
C1 for explaining that the back bearing differs by 180° (add 180°), not \(180 - 70\)
Worked solution: Going from \(B\) back to \(A\) is the opposite direction, which is 180° different from 070°.
(b)Answer: 250°
B1 for 250
Worked solution: \(070 + 180 = 250^\circ\)
Question 2Medium3 marks
The map shows the positions of the harbour (\(P\)) and the pier (\(Q\)). The map is accurately drawn. Use the scale bar: 1 cm on the map represents 1 km. A buoy, \(B\), is on a bearing of 150° from \(P\) and on a bearing of 205° from \(Q\).
[object Object]
(a) On the map, mark the position of \(B\) with a cross (×). Label it \(B\).[2]
(b) Use the map to find the distance of \(B\) from \(P\). Give your answer in km.[1]
Show the answer and mark scheme
(a)Answer: Lines drawn on a bearing of 150° from \(P\) and 205° from \(Q\); \(B\) marked where they cross (6.6 cm from \(P\) on the map).
M1 for a line from \(P\) on a bearing of 150° or from \(Q\) on a bearing of 205°, within ±2°
A1 for both lines drawn within ±2° and \(B\) marked at their intersection
Worked solution: Place the protractor on the north line at \(P\) and measure 150° clockwise; draw a long line. Do the same at \(Q\) for 205°. \(B\) is where the lines cross.
(b)Answer: 6.6 km (accept 6.1 to 7.3 km) km
B1 for 6.1 to 7.3 km (ft their \(B\): their length in cm × 1)
Worked solution: \(PB\) measures about 6.6 cm on the map, which represents \(6.6 \times 1 \approx 6.6\) km.
Question 3Hard6 marks
The map shows the positions of Elmsworth (\(P\)) and Faywick (\(Q\)). The map is accurately drawn. Use the scale bar: 1 cm on the map represents 2 km. A church tower, \(T\), is on a bearing of 075° from \(P\) and on a bearing of 355° from \(Q\).
[object Object]
(a) On the map, mark the position of \(T\) with a cross (×). Label it \(T\).[2]
(b) Use the map to find the distance of \(T\) from \(P\). Give your answer in km.[1]
(c) Use the map to find the distance of \(T\) from \(Q\). Give your answer in km.[1]
(d) Work out the bearing of \(P\) from \(T\).[2]
Show the answer and mark scheme
(a)Answer: Lines drawn on a bearing of 075° from \(P\) and 355° from \(Q\); \(T\) marked where they cross (5.5 cm from \(P\) on the map).
M1 for a line from \(P\) on a bearing of 075° or from \(Q\) on a bearing of 355°, within ±2°
A1 for both lines drawn within ±2° and \(T\) marked at their intersection
Worked solution: Place the protractor on the north line at \(P\) and measure 075° clockwise; draw a long line. Do the same at \(Q\) for 355°. \(T\) is where the lines cross.
(b)Answer: 11 km (accept 10.4 to 11.6 km) km
B1 for 10.4 to 11.6 km (ft their \(T\): their length in cm × 2)
Worked solution: \(PT\) measures about 5.5 cm on the map, which represents \(5.5 \times 2 \approx 11\) km.
(c)Answer: 7 km (accept 6.2 to 7.8 km) km
B1 for 6.2 to 7.8 km (ft their \(T\))
Worked solution: \(QT\) measures about 3.5 cm on the map, which represents about 7 km.
(d)Answer: 255°
M1 for \(75 + 180\) or for a correct method using the north line at \(T\) (e.g. alternate or co-interior angles)
A1 for 255°
Worked solution: The bearing of \(P\) from \(T\) is the back bearing: \(75 + 180 = 255\)°, so 255°.