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Calculator 1 mark
The diagram shows a harbour \(H\) and a boat \(B\).
\(B\) is west of \(H\).
State the bearing of \(B\) from \(H\).
Show answer
\(270^\circ\)A bearing is measured clockwise from north, so start facing north and turn clockwise until you are facing \(B\).
west is \(270^\circ\) clockwise from north.
A bearing is always written with three figures, so anything under 100 needs a leading zero.
The bearing of \(B\) from \(H\) is \(270^\circ\).
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Calculator 1 mark
The diagram shows a school \(S\) and a mast \(R\). The bearing of \(R\) from \(S\) is marked on it.
State the compass direction of \(R\) from \(S\).
Show answer
southThe arc starts at the north line and turns clockwise, so read it as a clockwise turn from north.
A quarter turn is \(90^\circ\) (east), a half turn is \(180^\circ\) (south) and three quarters is \(270^\circ\) (west). The in-between directions land on \(45^\circ\), \(135^\circ\), \(225^\circ\) and \(315^\circ\).
\(180^\circ\) clockwise from north is south.
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Either paper 5 marks
Here is a map of a town.
Each square of the grid is 1 cm by 1 cm.
Scale: 1 cm represents 136 m
North is at the top of the map.
(a)Which place is exactly south east of the station?
Circle your answer.
library, market, monument, school, town hall
(b)Circle the three-figure bearing of the town hall from the station.
000° 090° 180° 270°
(c)What is the distance, in metres, from the market to the station?
(d)Why might the shortest walking distance from the market to the station be greater than your answer to part (c)?
Show answer
(a) Monument
(b) 180°
(c) \(544\ \text{m}\)
(d) The answer to part (c) is the straight-line distance, but a walk has to follow roads and paths, which cannot go in a straight line between the two places, so the route is longer.Exactly south east of the station means the same number of centimetres south east on the grid, along a diagonal of the squares.
The monument is 4 cm east and 4 cm south of the station, so it is on that diagonal.
The town hall and the station are on the same vertical grid line, so the direction is due south.
Bearings are measured clockwise from north, so due south is \(180^\circ\).
The two places are on the same horizontal grid line, so count the centimetre gaps: \(4\) cm.
Each centimetre represents 136 m:
\(4 \times 136 = 544\)
The distance is \(544\) m.
The distance in part (c) is the straight line between the two places on the map, which is the shortest distance possible.
People walk along roads and paths, which bend and turn rather than going straight, so the walking route is longer.
- (c) M1counting \(4\) cm between them on the map
- (c) A1\(4 \times 136 = 544\) m
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Calculator 2 marks
The diagram shows two points \(A\) and \(B\), with a north line drawn at each of them.
The bearing of \(B\) from \(A\) is \(127^\circ\).
Work out the bearing of \(A\) from \(B\). This is the angle marked \(x\).
Show answer
\(307^\circ\)Both bearings describe the same straight line \(AB\), read from opposite ends.
Every north line points straight up the page, so the north line at \(A\) and the north line at \(B\) are parallel, and \(AB\) is a line crossing both of them. The angle \(AB\) makes with north at one end and the angle it makes with south at the other are alternate angles, so they are equal — and that is what makes the two bearings differ by exactly half a turn, \(180^\circ\).
\(127\) is less than 180, so adding a half turn keeps the answer under 360.
\(127 + 180 = 307\)
The bearing of \(A\) from \(B\) is \(307^\circ\).
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Calculator 2 marks
A coastguard at \(S\) can see a yacht at \(Y\) and a buoy at \(U\).
The bearing of \(Y\) from \(S\) is \(061^\circ\).
The bearing of \(U\) from \(S\) is \(152^\circ\).
Work out the size of the angle marked \(x\).
Show answer
\(91^\circ\)Both bearings are measured from the same north line, at \(S\), and both are measured clockwise from it. So the angle between the two lines is just the difference between the two bearings.
\(152 - 061 = 91\)
Angle \(x = 91^\circ\).
Note that this is an ordinary angle, not a bearing, so it is written \(91^\circ\) — no leading zeroes.
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Calculator 6 marks
Here is a map showing points \(P\) and \(Q\).
(a)Mateo wants to measure the bearing of \(P\) from \(Q\).
Mateo draws the line \(QP\) and a north line at \(Q\), then measures the angle clockwise from the north line to the line \(QP\). The angle is \(85^\circ\).
Mateo says that the bearing of \(P\) from \(Q\) is \(085^\circ\).
Is Mateo's method correct? Give a reason for your answer.
(b)On a different map, the bearing of \(D\) from \(C\) is \(315^\circ\).
Yara says, "\(D\) is south east of \(C\)."
Is Yara correct? Give a reason for your answer.
(c)A map of an island is drawn on a grid of squares of side \(2\) cm. An airport, \(E\), is marked on the map.
Scale: \(1\) cm represents \(14\) km.
A plane flies due south from the airport. On the map the coast is \(5\) squares due south of \(E\).
How far does the plane fly until it reaches the sea?
Show answer
(a) Yes: a bearing of P from Q is measured clockwise from the north line at Q, and that is what Mateo did
(b) No: a bearing of 315° points north west
(c) \(140\ \text{km}\)A bearing of \(P\) from \(Q\) starts at \(Q\), the place the bearing is measured from, and turns clockwise from the north line to the line \(QP\).
Mateo measured clockwise from the north line at \(Q\).
Answer: Yes, because a bearing of P from Q is measured clockwise from the north line at Q, and that is what Mateo did.
The compass points between north, east, south and west lie at \(045^\circ\), \(135^\circ\), \(225^\circ\) and \(315^\circ\), taken clockwise from north in that order: north east, south east, south west, north west.
A bearing of \(315^\circ\) is north west.
Yara says south east, so the answer is No.
Distance on the map: \(5 \times 2 = 10\) cm.
Each centimetre is \(14\) km, so \(10 \times 14 = 140\) km.
- (a) A1Yes
- (a) B1reason: the angle must be measured clockwise from north at Q
- (b) A1No
- (b) B1reason: 315° is north west, not south east
- (c) M1map distance \(10\) cm
- (c) A1distance \(140\) km
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Calculator 2 marks
A helicopter travels in a straight line from the hospital to the island on a bearing of \(117^\circ\).
It then travels straight back from the island to the hospital.
Calculate the bearing of the return journey.
Show answer
\(297^\circ\)There is no diagram, so sketch one: a north line at each place and a straight line joining them. The two north lines are parallel, so the outward and return bearings are half a turn apart.
Add \(180\), because a bearing has to stay between \(000^\circ\) and \(360^\circ\).
\(117 + 180 = 297\)
The return bearing is \(297^\circ\).
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Calculator 2 marks
Yusuf cycles from \(A\) to \(B\) on a bearing of \(319^\circ\).
At \(B\) he turns \(90^\circ\) anticlockwise and carries straight on.
Find the bearing he is cycling on now.
Show answer
\(229^\circ\)Bearings are measured clockwise, so turning clockwise makes the bearing bigger and turning anticlockwise makes it smaller. This turn is anticlockwise.
\(319 - 90 = 229\)
That is already between 000 and 360, so nothing more to do.
Yusuf is now cycling on a bearing of \(229^\circ\).
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Calculator 4 marks
A ferry sails from \(P\) to \(Q\) on a bearing of \(080^\circ\).
It then sails from \(Q\) to \(R\) on a bearing of \(146^\circ\).
(a)Find the size of angle \(PQR\), marked \(x\) on the diagram.
(b)Find the angle it turns through at \(Q\).
Show answer
(a) \(114^\circ\)
(b) \(66^\circ\)First find the bearing of \(P\) from \(Q\) — the direction it has come from. The north lines at \(P\) and \(Q\) are parallel, so that is half a turn from the first bearing:
\(080 + 180 = 260\)
Angle \(PQR\) is the angle between \(QP\) and \(QR\). Both of those are measured clockwise from the same north line at \(Q\), so subtract:
\(260 - 146 = 114\)
Angle \(PQR = 114^\circ\).
The turn is the change in the bearing it is travelling on:
\(146 - 080 = 66\)
It turns through \(66^\circ\).
The turn and angle \(PQR\) are not the same thing, and mixing them up is the commonest mistake here. If it carried straight on past \(Q\) it would be going in a straight line, so the turn and angle \(PQR\) sit on a straight line together and add to \(180^\circ\): \(66 + 114 = 180\).
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Calculator 3 marks
The diagram shows the route of a yacht. It sails from \(A\) to \(B\), and then from \(B\) to \(C\).
The bearing of \(B\) from \(A\) is \(058^\circ\), and angle \(ABC = 102^\circ\).
Calculate the bearing of \(C\) from \(B\).
Show answer
\(136^\circ\)Angle \(ABC\) is the angle inside the corner. What changes the bearing is the angle the yacht turns through, and the two sit on a straight line together, so they add to \(180^\circ\):
\(180 - 102 = 78\)
The route bends to the right — clockwise — so the bearing goes up by that much:
\(058 + 78 = 136\)
The bearing of \(C\) from \(B\) is \(136^\circ\).
The longer route gets there too, and is worth knowing as a check. The north lines at \(A\) and \(B\) are parallel, so the bearing of \(A\) from \(B\) is \(058 + 180 = 238\). Turning anticlockwise from \(BA\) through angle \(ABC\) reaches \(BC\): \(238 - 102 = 136\).