F.2 英文圖形題,20分

(a)
(i)
Draw DH perpendicular to OA and cut OA produce at H.

AB = CD = 120 cm
BC = DA = 45 cm
∠DAH = ∠ABO = θ

In ΔDAH :
sinθ = DH/DA
sin60° = DH/(45 cm)
DH = 45 sin60° cm

Distance between point D and the wall OA
= DH
= 45 sin60° cm
= 39.0 cm

(ii)
In ΔABO :
sinθ = OA/AB
sin60° = OA/(120 cm)
OA = 120 sin60° cm

In ΔDAH :
cosθ = AH/DA
cos60° = AH/(45 cm)
AH = 45 cos60° cm

Distance between point D and the wall OB
= OA + AH
= (120 sin60° + 45 cos60°) cm
= 126 cm

(b)
√(120² + 45²) = 15√73

Distance between point D and the wall OB (in cm)
= 120 sinθ + 45 cosθ
= 15√73[(120/15√73)sinθ + (45/15√73)cosθ]
= 15√73(cos20.6° sinθ + sin20.6 cosθ)
= 15√73 x sin(20.6° + θ)

The distance is the longest when sin(20.6° + θ) = 1
20.6° + θ = 90°
θ = 69.4°

The longest distance
= 15√73 cm
= 128 cm

√(120² + 45²) = 15√73係咩？