✔ 最佳答案
14.
(a) (i)
PQ = 12 x 3 = 36 km
(a) (ii)
PR² = PQ² + QR² - 2 PQ QR cosQ
PR² = 36² + 51² - 2 x 36 x 51 x cos40° km²
PR = 32.9 km
(b)
Average speed of ship M
= 32.9/2 km/h
= 16.5 km/h
(c)
sinR / PQ = sinQ / PR
sinR / 36 = sin40° / 32.9
R = 44.7°
Bearing of R from P = S 44.7° E
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15.
(a)
The bearing of P from aeroplane B
= S [180° - (285° - 180°)] E
= S 75° E
(b)
In DLBP :
∠PLB = 230° - 155° = 75°
∠BPL = 180° + 155° - 285° = 50°
∠PBL = 180° - (75° + 50°) = 55°
sin∠BPL / BL = sinPLB / BP (sine law)
sin50° / BL = sin 75° / (60 km)
BL = 60 sin50° / sin75° km = 47.6 km
(c)
Denote H as the hangar.
∠LBH = 230° - 180° = 50°
In DBLH :
cos∠LBH = BH / BL
cos50° = BH / (47.6 km)
Distance that the aeroplane have to move BH
= 47.6 cos50° km
= 30.6 km