Write down the three-figure bearing of each direction.
(a) East [1]
(b) South-west [1]
- (a) $090°$ (accept 090)B1
- (b) $225°$B1
- Answer: (a) $090°$ (b) $225°$
- Examiner tip: Bearings always have three figures — 90° should be written 090°.
(a) The bearing of B from A is $070°$.
Work out the bearing of A from B. [2]
(b) The bearing of A from C is $310°$.
Work out the bearing of C from A. [2]
- (a) $70 + 180$ (or a correct sketch with parallel north lines and $180 - 70 = 110$ shown)M1
- (a) $250°$A1
- (b) $310 - 180$M1
- (b) $130°$A1
- Answer: (a) $250°$ (b) $130°$
- Examiner tip: For a back bearing add 180° if the bearing is less than 180°, otherwise subtract 180°.
A boat sails from P to Q on a bearing of $115°$.
It then sails from Q to R on a bearing of $220°$.
Work out the size of angle PQR. [3]
- finds the bearing of P from Q, $115 + 180 = 295$, or angle between north at Q and QP $= 180 - 115 = 65$ (co-interior angles)M1
- complete method, e.g. $295 - 220$ or $(360 - 220) - 65$ or $180 - 65 - (220-180)$M1
- $75°$A1
- Answer: $75°$
- Examiner tip: Draw a north line at Q and mark the 220° turning clockwise from it — most errors come from sketching R on the wrong side.
B is 8 km due east of A.
C is 5 km due north of B.
Work out the bearing of A from C.
Give your answer to the nearest degree. [4]
- a correct trig statement for an angle in triangle ABC, e.g. $\tan(\angle ACB) = \dfrac{8}{5}$ or $\tan(\angle CAB) = \dfrac{5}{8}$M1
- $\angle ACB = 57.99...°$ (or $\angle CAB = 32.00...°$)A1
- $180 + 57.99...$ (or $270 - 32.00...$)M1
- $238°$A1
- Answer: $238°$
- Examiner tip: Bearing 'of A from C' means stand at C and face north — the angle is measured from C, not from A.
A ship sails 12 km from port P to a buoy A on a bearing of $060°$.
It then sails 9 km from A to a lighthouse B on a bearing of $150°$.
(a) Show that angle PAB $= 90°$. [1]
(b) Work out the distance PB. [2]
(c) Work out the bearing of B from P. Give your answer to the nearest degree. [2]
- (a) e.g. bearing of P from A is $060 + 180 = 240$ and $240 - 150 = 90$ (or angle between north at A and AP is $180-60=120$ by co-interior angles, so $\angle PAB = 360 - 120 - 150 = 90$)B1
- (b) $\sqrt{12^2 + 9^2}$M1
- (b) $15$ kmA1
- (c) $\tan^{-1}\left(\dfrac{9}{12}\right) = 36.86...°$ (or $\sin^{-1}(9/15)$ etc.)M1
- (c) $097°$ (from $60 + 36.86... = 96.86...$)A1
- Answer: (b) 15 km (c) $097°$
- Examiner tip: The angle you find with trig is inside the triangle — you must add it to the 60° to get a bearing from north.
A walker goes 7 km from P to Q on a bearing of $035°$.
She then walks 10 km from Q to R on a bearing of $110°$.
(a) Work out the distance PR. Give your answer correct to 3 significant figures. [3]
(b) Work out the bearing of R from P. Give your answer to the nearest degree. [3]
- (a) finds angle PQR $= 105°$, e.g. bearing of P from Q is $035 + 180 = 215$ and $215 - 110 = 105$ (or $360 - (180-35) - 110$)M1
- (a) $PR^2 = 7^2 + 10^2 - 2\times7\times10\times\cos 105°$M1
- (a) $13.6$ km (from $13.61...$; accept answers rounding to 13.6, e.g. 13.61)A1
- (b) correct method for angle QPR, e.g. $\dfrac{\sin QPR}{10} = \dfrac{\sin 105°}{13.61...}$ or $\cos QPR = \dfrac{7^2 + 13.61...^2 - 10^2}{2\times7\times13.61...}$ (ft their PR)M1
- (b) angle QPR $= 45.2...°$ (accept 45.2 to 45.3, e.g. 45.25 from using PR = 13.6)A1
- (b) $080°$ (from $35 + 45.21... = 80.21...$; accept 80°)A1
- Answer: (a) 13.6 km (b) $080°$
- Examiner tip: Use the unrounded value of PR (13.610…) in the sine rule, and remember angle QPR is inside the triangle — add it to the 35° to get the bearing.
(a) Write down the three-figure bearing of north-west. [1]
(b) Write down the compass direction that is on a bearing of $270°$. [1]
(c) The diagram shows the positions of two points, P and Q.
Work out the bearing of Q from P. [1]
- (a) $315°$ (accept 315)B1
- (b) West (accept W)B1
- (c) $325°$ (from $360 - 35$; do not accept $035°$ or $35°$)B1
- Answer: (a) $315°$ (b) West (c) $325°$
- Examiner tip: Bearings are always measured clockwise from north, so an angle marked anticlockwise from north must be taken away from 360°.
The diagram shows the positions of two villages, A and B, on a map.
The scale of the map is 1 cm represents 5 km.
On the map, the distance from A to B is 4.8 cm.
(a) Work out the real distance from A to B. [2]
(b) Write down the bearing of B from A. [1]
- (a) $4.8 \times 5$M1
- (a) $24$ kmA1
- (b) $072°$ (accept $72°$)B1
- Answer: (a) 24 km (b) $072°$
- Examiner tip: Multiply the map length by the scale (do not divide), and write a bearing under 100° with a leading zero, e.g. 072°.
A ship leaves port P at 09:00.
It sails on a bearing of $125°$ at a constant speed of 16 km/h.
At 11:30 the ship stops at point S.
(a) Work out the distance PS. [2]
Ama draws the route PS on a map with scale 1 : 500 000.
(b) Work out the length, in centimetres, of the line PS on her map. [2]
(c) The ship then sails directly back to port P.
Write down the bearing it must sail on. [1]
- (a) $16 \times 2.5$ (accept $16 \times 2\tfrac{1}{2}$; $16 \times 2.3$ is M0)M1
- (a) $40$ kmA1
- (b) converts to cm and divides by the scale, e.g. $40 \times 100\,000 \div 500\,000$ or $4\,000\,000 \div 500\,000$ (ft their (a)); or uses 1 cm represents 5 km, so $40 \div 5$M1
- (b) $8$ cm (ft their (a))A1
- (c) $305°$B1
- Answer: (a) 40 km (b) 8 cm (c) $305°$
- Examiner tip: 09:00 to 11:30 is 2.5 hours, not 2.3 hours; and 1 km = 100 000 cm when you change a ratio scale into km.
The bearing of Y from X is $245°$.
Tom says, “The bearing of X from Y is $245° + 180° = 425°$.”
(a) Explain why Tom’s answer cannot be correct. [1]
(b) Work out the bearing of X from Y. [1]
The bearing of Q from P is $x°$, where $x < 180$.
The bearing of P from Q is $(5x - 20)°$.
(c) Work out the value of $x$. [3]
- (a) bearings must be less than $360°$ (or he should have subtracted $180°$ because $245 > 180$; accept $425 - 360 = 65$ shown with explanation)C1
- (b) $065°$ (from $245 - 180$; accept $65°$)B1
- (c) forms a correct equation, $x + 180 = 5x - 20$M1
- (c) correct rearrangement, e.g. $4x = 200$M1
- (c) $x = 50$A1
- Answer: (a) a bearing must be less than 360° (b) $065°$ (c) $x = 50$
- Examiner tip: Back bearing: add 180° if the bearing is under 180°, subtract 180° if it is 180° or more, so the answer always stays between 000° and 360°.
The diagram shows three buoys, A, B and C.
B is due east of A.
The bearing of C from A is $050°$.
The bearing of C from B is $335°$.
(a) Work out the size of angle ACB. [3]
(b) Work out the bearing of A from C. [1]
- (a) angle CAB $= 90 - 50 = 40°$ (may be on diagram)B1
- (a) angle ABC found, e.g. $90 - (360 - 335) = 65°$ or $335 - 270 = 65°$ (may be on diagram)M1
- (a) $75°$ (from $180 - 40 - 65$)A1
- (b) $230°$ (from $050 + 180$)B1
- Answer: (a) $75°$ (b) $230°$
- Examiner tip: Because B is due east of A, line AB points along 090°/270°, so each angle inside the triangle is found from the gap between the bearing and 90° or 270°.
A yacht race follows a course in the shape of a regular pentagon ABCDE.
The yachts sail clockwise around the course, from A to B to C to D to E and back to A.
The bearing of B from A is $030°$.
(a) Show that angle ABC $= 108°$. [2]
(b) Work out the bearing of C from B. [2]
(c) Work out the bearing of E from A. [2]
- (a) $(5 - 2) \times 180 = 540$ (or exterior angle $360 \div 5 = 72$)M1
- (a) $540 \div 5 = 108$ (or $180 - 72 = 108$) with no errors seenA1
- (b) bearing of A from B $= 030 + 180 = 210$ (or uses co-interior angle $180 - 30 = 150$ at B, or turns through exterior angle $030 + 72$)M1
- (b) $102°$ (e.g. $210 - 108$ or $360 - 150 - 108$)A1
- (c) $030 + 108$ (or bearings of the legs $030 + 72 \times 4 = 318$ for E to A, then $318 - 180$)M1
- (c) $138°$A1
- Answer: (a) shown: $540 \div 5 = 108$ (b) $102°$ (c) $138°$
- Examiner tip: At B the bearing back to A is 210°; the angle ABC = 108° is measured anticlockwise from BA, so the bearing of C from B is 210° − 108° = 102° (not 108°).
A helicopter flies 36 km from H to J on a bearing of $132°$.
It then flies 20 km due west from J to L.
(a) Work out how far south of H the point J is.
Give your answer correct to 3 significant figures. [2]
(b) Work out the bearing of L from H.
Give your answer to the nearest degree. [3]
- (a) $36\cos 48°$ (or $36\sin 42°$, or $36 \cos 132°$ with sign ignored)M1
- (a) $24.1$ km (from $24.08...$)A1
- (b) distance of J east of H, $36\sin 48° = 26.75...$ (or $36\cos 42°$), and $26.75... - 20 = 6.75...$M1
- (b) $\tan^{-1}\left(\dfrac{6.75...}{24.08...}\right) = 15.66...°$ (or $\tan^{-1}\left(\dfrac{24.08...}{6.75...}\right) = 74.3...°$) (ft their (a))M1
- (b) $164°$ (from $180 - 15.66... = 164.3...$ or $90 + 74.3...$)A1
- Answer: (a) 24.1 km (b) $164°$
- Examiner tip: The angle between the flight path and due south is 180° − 132° = 48°, so the southward distance uses cos 48°, not cos 132° or sin 132°.
Two coastguard stations, A and B, are on a straight east–west coastline.
B is 14 km due east of A.
A ship, S, is on a bearing of $038°$ from A and on a bearing of $312°$ from B.
(a) Work out the distance AS.
Give your answer correct to 3 significant figures. [3]
(b) Work out the shortest distance from the ship to the line AB.
Give your answer correct to 3 significant figures. [2]
- (a) finds two angles of triangle ABS: angle SAB $= 52°$, angle SBA $= 42°$ (from $312 - 270$) and so angle ASB $= 86°$M1
- (a) $\dfrac{AS}{\sin 42°} = \dfrac{14}{\sin 86°}$ (ft their angles if two correct angles used)M1
- (a) $9.39$ km (from $9.390...$)A1
- (b) $9.390...\times \sin 52°$ (ft their AS; or $BS \times \sin 42°$ with $BS = 11.05...$)M1
- (b) $7.40$ km (accept $7.4$; from $7.399...$)A1
- Answer: (a) 9.39 km (b) 7.40 km
- Examiner tip: Turn the bearings into angles inside the triangle first (52° at A, 42° at B), then use the sine rule with the side opposite the 86° angle.
The diagram shows the positions of three towns, P, Q and R.
PQ = 18 km, QR = 22 km and PR = 25 km.
The bearing of Q from P is $070°$.
R is south-east of the line PQ, as shown.
(a) Work out the bearing of R from P.
Give your answer to the nearest degree. [4]
(b) Write down the bearing of P from R. [1]
- (a) $\cos QPR = \dfrac{18^2 + 25^2 - 22^2}{2 \times 18 \times 25}$ (or $22^2 = 18^2 + 25^2 - 2\times18\times25\cos QPR$)M1
- (a) angle QPR $= 58.89...°$ (accept 58.9; $\cos QPR = \tfrac{465}{900}$)A1
- (a) $070 + 58.89...$M1
- (a) $129°$ (from $128.89...$)A1
- (b) $309°$ (ft their (a) $+ 180$)B1
- Answer: (a) $129°$ (b) $309°$
- Examiner tip: Find angle QPR with the cosine rule first; the bearing of R is then the bearing of Q plus that angle, because R lies clockwise from Q as seen from P.
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