急 三角比問題(應用) 20分

3小時後A離O 60km, B離O 45km

A和B之間的夾角 = 90 + 50 = 140

用cosine law,3小時後A和B的距離

√ (60^2 + 45^2 - 2(60)(45)cos140)

= 98.8 km

A


O


B C

After 3 hours, A moves 60 km from O and B moves 45 km from O
AO = 60 km OB = 45 km
Angle COB = 50度
Line AOC is perpendicular to line BC
Consider triangle OBC
Cos angle COB = OC/OB
Cos angle 50度 = OC/45 km
OC = 45 km cos 50° = 45 km(0.6428) =28.9254 km
sin angle COB = BC/OB
sin 50度 = BC/45 km
BC = 45 km sin 50° = 45 km (0.7660) = 34.4720 km
AC = OA + OC = 60 km + 28.9254 km =88.9254 km
Now consider triangle ABC
AB^2 = BC^2 + AC^2 = (34.4720 km )^2 + (88.9254 km)^2 (Pythagorean Theorem)
AB^2 = 1188.32 + 7907.73 km^2 = 9096.05 km^2
AB = square root of 9096.05 km^2 = (9096.05)^0.5 km = 95.37 km
A和B的距離 = 95.37 km

Using cosine Law to check the answer:
√ (60^2 + 45^2 - 2(60)(45)cos130°)
= √[ (3600 + 2025 -2(60)(45) (-0.6428)]
= √ (3600 + 2025 +3471.05)
= √9096.05
= 95.37 km

To: 回答者：myisland8132 ( 知識長 )
No offence:
A和B之間的夾角不是140,
應該是 130度
You get your bearing wrong.
Bearing 是S50度W (making 40度 with the line pointing West)