F.4 數學...(3D)

a) By sine law,
PR / sin β = c / sin(180 - (α + β)
PR = c sin β / (sinα + β)
tan θ = h / PR
tan θ = h sin (α + β) / c sin β
h = c tanθ sin β / sin (α + β)

b)(i) h = c tanθ sin β / sin (α + β)
h/c = tan 40°sin 46° / sin (54° + 46°)
h/c = 0.613

(ii) By sine law,
QR / sin α = c / sin(α + β)
QR = c sin 54° / sin 100°
tan ∠BQR = h / QR
= h/c ( sin 100° / sin 54°)
∠BQR = 36.7°

(iii) By cosine law,
cos α = (PR^2 + PM^2 - MR^2) / (2)(PR)(PM)
cos 54° = ((c sin 46° / sin 100°)^2 + (c / 2)^2 - MR^2) / 2(c sin 46° / sin 100°)(c / 2)
MR = 0.595c
tan ∠BMR = h/ 0.595c
= (h/c)(1 / 0.595)
∠BMR = 45.8°

PR = 0.7304c , MR = 0.595c , PM = 0.5c
By cosine law,
cos ∠PRM = PR^2 + MR^2 - PM^2 / 2(PR)(PM)
∠PRM = 42.8°
∠PRM = 180° - 54° - 42.8° = 83.2°

The bearing is N 83.2° E

(a) angle PQR = 180 - α - β
By sine law
PR / sin β = c / sin (180 - α - β)
PR = c sin β / sin (180 - α - β)
PR = h / tan θ
h / tan θ = c sin β / sin (α + β)
h = c tanθ sin β / sin (α + β)
(b)(i)
α=54, β= 46,θ =40
from result (a)
h/c = tanθ sin β / sin (α + β)
= tan 40 sin 46 / sin (46+54)
= 0.613
(ii)
QR / sinα = c / sin (α + β)
h/QR = tan(angle BQR)
h/ (c sinα / sin(α + β) = tan (angle BQR)
tan (angle BQR) = 0.613 sin (46 + 54) / sin 40
= 0.939
angle BQR = arc tan 0.939 = 43.2
(iii)
PR = h / tanθ, PM = c/2 = h/(2 x 0.613) = 0.816 h
By cosine law:
RM^2 = PR^2 + PM^2 - 2(PM)(PR)cos α
= (h/ tan40)^2 + (0.816h)^2 - 2(h/tan40)(0.816h)cos54
= 1.42h^2 +0.665h^2 - 1.1429h^2
= 0.943h^2
RM = 0.971 h
tan (angle BMR) = h / RM
= h / (0.971h)
= 1.03
angle BMR = arc tan (1.03) = 45.8

2008-03-09 23:33:10 補充：
Further to question (3)
MQ = PM =0.816h
By sine law
MQ / sin (angle MRQ) = MR / sin β
0.816h / sin (angle MRQ) = 0.971h / sin 46°
angle MRQ = 37.2°
Bearing from B to Q is S 46°W
Bearing from B to M is S46°W + 37.2° = S83.2°W