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.)
