O lvl E Maths: Trigonometry Bearings

In the diagram, the bearing of A from B is 320° and the bearing of B from C is 240°. Given that AB = BC = 10 km, calculate
(i) ∡BCA,
(ii) the bearing of C from A.

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Answer:

(i) ∡QCB = 240° - 180° = 60°
∡NBC = 60° (alternate angles)
∡NBA = 360° - 320° = 40° (angles at a point)
∡ABC = 40° + 160° = 100°
∡BCA = (180° - 100°) / 2 = 40° (angles in an isosceles triangle)


(ii) ∡ACQ = 40° + 60° = 100°
Bearing of C from A = 100° (alternate angles)

O lvl E Maths: Trigonometry --> Bearings

RI sec 3 1997 EOY P2 Q17

A straight road joins 3 locations A, B and C. The road runs in a direction of 042° from A. A man at P is 75 metres due south of A. A tower at C has angle of elevation of 45° from the man. Given that PC = 150 m and AB = 65 m. Calculate

(i) the bearing of P from C.
(ii) the distance between P and B.
(iii) the angle of elevation of the tower from A.
(iv) the area of ΔPBC

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Answer:

∡ACQ = 42° (alternate angles)
∡CAP = 180° - 42° = 138°

(i) Using sine rule for ΔACP,
sin ∡ACP / 75 = sin 138° / 150
sin ∡ACP = sin 138° / 150 * 75 = 0.3345653
∡ACP = 19.546°
Since ∡ACQ = 42°,
Bearings = 180° + 42° - 19.546° = 202.5°

(ii) Using cosine rule,
BP² = 75² + 65² - 2 (75)(65) cos 138°
BP ≈ 130.75 m
BP = 131 m (3 s.f.)

(iii)

tan 45° = h / PC
h = 150 m

From the first diagram, using sine rule,
AC / sin 202.454° = 150 / sin 138°
AC = 85.62 m

tan θ = h / 85.62 = 150 / 85.62
θ = 60.3°


(iv) BC = 85.62 - 62 = 20.62 m
Area of ΔPBC = ½ * 20.62 * 150 * sin 19.546°
= 517 m² (3 s.f.)

O lvl E Maths: Trigonometry --> Bearings

Question from http://www.sgforums.com/forums/2297/topics/331427

A point B is 500 m due North of a point A. Rahmat walks from A on a bearing of 325°.

Find Rahmat's distance from A when he is

(i) due West of B

(ii) equidistant from A and B

(iii) closest to B

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Answer:

(i) bearings of 325° = 35° anti-clockwise as shown:

Hence, distance from A = 500 / cos 35°
= 610.4 m


(ii) Diagram is as shown:



Distance from A = 250 / cos 35°
= 305.2 m


(iii) Diagram is as shown:



Distance from A = 500 * cos 35°
= 409.6 m

O lvl E Maths: Trigonometry --> Bearings

P, Q and R are 3 points on the same horizontal ground. The bearing of Q from P is 030° and PQ is 1200 m apart and hot air balloon is spotted vertically above Q at an angle of elevation of 21° from P. The balloon is drifting at a constant height towards the east at a steady speed of 12 km/h. Five minutes after passing Q , it is observed to be directly above R . Find

1) The distance between P and R
2) The bearing of R from p
3) the angle of depression of P from the balloon when it is directly above R

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Answer:

1) After 5 minutes, the balloon would have travelled east at 12 km/h× 5/60 h = 1 km

So if we draw out the points P, Q and R, it will look like the following:



So angle PQR = 90° + 30° = 120°

Using cosine rule,
PR² = PQ² + QR² - 2 × (PQ)(QR)(cos 120°)
PR² = 1.2² + 1² - 2 × (1.2)(1)(-½)
PR² = 3.64
PR = 1.908 km


2) Using sine rule,
(sin ∡QPR ) / 1 = (sin ∡PQR) / 1.908
sin ∡QPR = sin 120 / 1.908
sin ∡QPR = 0.45389
∡QPR = 27°

Hence, bearings of R from P = 030 + 027 = 057


3) The diagrams can be redrawn to have a side view so as to see it clearer



Vertical height of balloon = 1.2 tan 21° = 0.4606 km

Because it is at a constant height, the following diagram is what happened when it reaches R:



Thus,
tan θ = 0.4606 / PR
tan θ = 0.4606 / 1.908
θ = 13.6°

Angle of depression of P from the balloon when it is directly above R = 13.6°