StudentAidUnit 7: Trigonometry — Exercise 7 4
10th Class Mathematics · Unit 7: Trigonometry
7.4.1.(i) Find the bearing of point P, where the line OP is 40° east of south. (ii) Find the bearing of point P, where the line OP is 20° north of west. (iii) Find the bearing of point P, where the line OP is 50° south of west. (iv) Find the bearing of point P, where the line OP is 75° north of east.
Formula
(i)
Bearing=180^circ-40^circ
Result
(i)
Bearing of P=140^circ
Formula
(ii)
Bearing=270^circ+20^circ
Result
(ii)
Bearing of P=290^circ
Formula
(iii)
Bearing=270^circ-50^circ
Result
(iii)
Bearing of P=220^circ
Formula
(iv)
Bearing=90^circ-75^circ
Result
(iv)
Bearing of P=15^circ
7.4.2.(i) The diagram shows the positions of two ships P and Q, with the line QP making a 60° angle with north at Q. Find the bearing of ship P from ship Q. (ii) The diagram shows the positions of two ships P and Q, with the line QP making a 60° angle with north at Q. Find the bearing of ship Q from ship P.
Given
(i)
The line QP is 60^circ west of North
Result
(i)
Bearing=360^circ-60^circ=300^circ
Given
(ii)
The line PQ is 60^circ east of South
Result
(ii)
Bearing=180^circ-60^circ=120^circ
7.4.3.(i) The diagram shows 3 places A, B and C, with angle ACB split into 50° (toward A) and 40° (toward B) from north at C. Find the bearing of A from C. (ii) The diagram shows 3 places A, B and C, with angle ACB split into 50° (toward A) and 40° (toward B) from north at C. Find the bearing of C from A. (iii) The diagram shows 3 places A, B and C, with angle ACB split into 50° (toward A) and 40° (toward B) from north at C. Find the bearing of C from B. (iv) The diagram shows 3 places A, B and C, with angle ACB split into 50° (toward A) and 40° (toward B) from north at C. Find the bearing of B from C.
Result
(i)
Bearing of A from C=360^circ-50^circ=310^circ
(ii) As printed
Bearing of C from A=180^circ+50^circ=230^circ
The bearing of A from C is 360-50=310° (N50°W). The reverse bearing is 310-180=130° (S50°E), not 180+50=230°; the wrong reciprocal formula was applied.
(iii)
Bearing of C from B=180^circ+40^circ=220^circ
(iv) As printed
Bearing of B from C=360^circ-40^circ=320^circ
B is 40° east of north at C, so the bearing of B from C is simply 0+40=40° (N40°E), not 360-40=320°. This also contradicts part (iii), whose reciprocal (220-180=40°) confirms 40° is correct.
7.4.4.(i) Abdul Hadi walks 100 m North and then 300 m East. How far is he from his starting position? (ii) Abdul Hadi walks 100 m North and then 300 m East. On what bearing should he walk to get back to his starting position?
Formula
(i) Pythagoras Theorem
d2=1002+3002=100000
Result
(i)
d=sqrt{100000}=100sqrt{10} m (approx.)
Given
(ii)
From P to O: Westward=300 m, Southward=100 m
(ii)
tanθ=100300=frac13 ⇒ θ=tan-1left(frac13right)=18.43^circ
Result
(ii)
Bearing=270^circ-18.43^circ=251.57^circ
7.4.5.Three ships L, M, N are in the position shown in the diagram (LM = 12 km, LN = 14 km, MN = 15 km). Ship M is North East of ship N. Find the bearing of L from M.
7.4.6.A ship sails 20 km on a bearing of 120°. How far to the South has the ship moved from its original position?
Given
Bearing 120^circ is 30^circ South of East
Formula
Southward component=20sin 30^circ
Result
=20 × frac12=10 km
7.4.7.Fatima walked South for 5.5 km and then turned West for 1.3 km. Calculate Huria's bearing from her starting point.
Given
Southward=5.5 km, Westward=1.3 km
tanθ=1.35.5=1355 ⇒ θ=tan-1left(1355right)=13.34^circ
Result
Bearing=180^circ+13.34^circ=193.34^circ
7.4.8.An aircraft flies 150 km east, then 100 km northeast (45° from East). What is the total displacement?
Given
First displacement: 150 km East; Second: 100 km at 45^circ
Working
Components of 2nd leg
East=100cos45^circ=50sqrt2, North=100sin45^circ=50sqrt2
Total East=150+50sqrt2, Total North=50sqrt2
Result
R=sqrt{(150+50sqrt2)2+(50sqrt2)2} ≈ 231.91 km
7.4.9.A ship sails 100 km on a bearing of 045°, then changes course and sails 120 km on a bearing of 135°. Find the distance between the starting point and the final position.
Working
First leg (100 km, 045°)
E1=100sin45^circ=50sqrt2, N1=100cos45^circ=50sqrt2
Second leg (120 km, 135°)
E2=120sin135^circ=60sqrt2, N2=120cos135^circ=-60sqrt2
Net Easting=110sqrt2, Net Northing=-10sqrt2
Result
R=sqrt{(110sqrt2)2+(10sqrt2)2}=sqrt{24400}=20sqrt{61} km (approx.)