1)
A man walks 400m from O to A on a bearing of 124度 and then 700m from A to B on a bearing of 25 .find the distance and the bearing of the final position from his starting point
2)
In Fig.3 ,A,B,C are 3 cities such that AB=80km,BC=120km,AC=100km and the bearing of B from A is N51度E .Find the bearing of C from A andB
Problems in 2 dimensions
2009-05-26 4:49 am
回答 (2)
2009-05-26 8:55 pm
✔ 最佳答案
1. Angle OAB = 25° + [ 90° - ( 124° - 90° ) ] = 81°OB^2 = 400^2 + 700^2 – 2 X 400 X 700 X cos81°
= 562396.6996
OB = 749.93
The distance of the final position from his starting point is 749.93 m
700 / sin (angle AOB) = 749.93 / sin 81°
sin (angle AOB) = 0.9219
angle AOB = 67.21°
The bearing the final position from his starting point is 124° - 67.21° = 56.79°
2. 120^2 = 80^2 + 100^2 – 2 X 80 X 100 X cos (angle BAC)
angle BAC = 82.82°
The bearing of C from A is S(180° - 51° - 82.82°)E i.e. S46.18°E
100^2 = 80^2 + 120^2 – 2 X 80 X 120 X cos (angle ABC)
angle ABC = 55.77°
The bearing of C from B is S(55.77° - 51°)E i.e. S4.77°E
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