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4.16F Bearings: Foundation Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write down the bearing of B from A.
A and B are two points. The direction from A to B is due east.

(Total for Question 1 is 1 mark)

Write down the bearing of C from A.
The direction from A to C is due south.

(Total for Question 2 is 1 mark)

A boat sails from harbour H to point P on a bearing of 045 degrees.
Work out the bearing of H from P.

(Total for Question 3 is 2 marks)

On a map, a path from X to Y has a bearing of 300 degrees.
Work out the bearing of X from Y.

(Total for Question 4 is 2 marks)

A walker stands at point W. The bearing of tree A from W is 072 degrees. The bearing of tree B from W is 162 degrees.
Work out the angle AWB between the two trees, as seen from W.

(Total for Question 5 is 3 marks)

Town P is due west of town Q. A plane leaves P and flies to R on a bearing of 030 degrees.
Work out (i) the bearing of Q from P, and (ii) the bearing of P from R. Show your working for both.

(Total for Question 6 is 4 marks)

On a city map, north is 000 degrees. From station T, the bearing to bus stop U is 255 degrees. From station T, the bearing to bus stop V is 345 degrees.
Work out (i) the angle UTV between the two bus stops, as seen from T, and (ii) the bearing of T from V.

(Total for Question 7 is 4 marks)

A lifeboat leaves harbour H and travels on a bearing of 015 degrees for 6 km to reach point L. At L it turns clockwise through a right angle (so its bearing increases by 90 degrees) and travels for 4 km to reach point M.
Work out the bearing of M from H. Give your answer to the nearest degree. Show your working clearly.

(Total for Question 8 is 5 marks)

A surveyor stands at point S. From S the bearing to monument A is 052 degrees and the bearing to monument B is 140 degrees. The surveyor is standing exactly the same distance from each monument (SA = SB).
Work out the bearing of B from A. Show your full reasoning and give the answer to the nearest degree.

(Total for Question 9 is 8 marks)

Two yachts leave harbour Y at the same time. Yacht 1 sails to buoy B on a bearing of 075 degrees for 8 km. Yacht 2 sails to buoy C on a bearing of 165 degrees for 8 km (the same distance).
Work out the bearing of B from C. You must show all steps and explain how you used the information. Give your answer to the nearest degree.

(Total for Question 10 is 10 marks)

Mark scheme · 4.16F Bearings: Foundation Tier Practice

Question 1

B1 090 degrees cao
Answer: 090 degrees

Question 2

B1 180 degrees cao
Answer: 180 degrees

Question 3

M1 since the bearing H->P is less than 180, add 180: 045 + 180
A1 225 degrees cao
Answer: 225 degrees

Question 4

M1 since the bearing X->Y is 180 or more, subtract 180: 300 - 180
A1 120 degrees cao
Answer: 120 degrees

Question 5

M1 recognise angle AWB is the difference between the two bearings measured from the same point W
A1 162 - 072 = 90 degrees cao
B1 clear final statement that angle AWB = 90 degrees cao
Answer: 90 degrees

Question 6

M1 recognise 'P is due west of Q' means Q is due east of P
A1 (i) 090 degrees cao
M1 back bearing of P from R: since the bearing P->R is less than 180, add 180: 030 + 180
A1 (ii) 210 degrees cao
Answer: (i) 090 degrees (ii) 210 degrees

Question 7

M1 recognise angle UTV is the difference between the two bearings measured from the same point T: 345 - 255
A1 (i) 090 degrees cao
M1 back bearing of T from V: since the bearing T->V is 180 or more, subtract 180: 345 - 180
A1 (ii) 165 degrees cao
Answer: (i) 090 degrees (ii) 165 degrees

Question 8

M1 recognise triangle HLM is right-angled at L (the turn is 90 degrees) with HL = 6 km, LM = 4 km
M1 use trigonometry: tan(angle LHM) = LM / HL = 4/6, so angle LHM = arctan(4/6)
A1 angle LHM = 33.7 degrees (1 dp)
B1 bearing of M from H = bearing of L from H + angle LHM = 015 + 33.7 = 048.7 cao
P1 communication: working shows the right-angled triangle, the trig ratio used, and the final bearing rounded to the nearest degree, 049 degrees
Answer: 049 degrees (to the nearest degree).

Question 9

M1 state relative bearings from S: S->A = 052, S->B = 140; find angle ASB = 140 - 52 = 88 degrees (method)
M1 recognise triangle SAB is isosceles (SA = SB) so the base angles are equal: base angle = (180 - 88)/2 = 46 degrees
M1 find the back bearing of S from A: 052 + 180 = 232 degrees
M1 combine: bearing A->B = bearing A->S - base angle = 232 - 46 (method)
A1 186 degrees cao
B1 clear reasoning linking the isosceles triangle, the back bearing and the subtraction of the base angle
P1 presentation: working clearly shows angle ASB, the isosceles base angle, and the back bearing calculation
A1 final answer 186 degrees cao
Answer: 186 degrees

Question 10

M1 find angle BYC = 165 - 075 = 90 degrees (method)
M1 recognise triangle YBC is isosceles (YB = YC = 8 km) so the base angles are equal: base angle = (180 - 90)/2 = 45 degrees
M1 find the back bearing of Y from C: 165 + 180 = 345 degrees (method)
M1 combine: bearing of B from C = bearing C->Y + base angle = 345 + 45 (method)
A1 390 - 360 = 030 degrees cao
B1 clear reasoning linking the isosceles triangle, the back bearing and the addition of the base angle
P1 presentation: steps shown clearly (angle BYC, base angle, back bearing, final bearing)
M1 alternative correct method: use a scale drawing or coordinate check to confirm B is on a bearing of 030 from C (method)
A1 final answer 030 degrees cao
C1 full correct solution with explanation and check (e.g. confirming triangle YBC is a right isosceles triangle, since angle BYC = 90)
Answer: 030 degrees

Bearings: Foundation Tier Practice - Worksheets, Questions and Revision