MATH Q

👨🏻‍🏫 學術台數學

用戶124294

2008-06-25 4:13 pm

2 ships A and B leave a port at the same time. Ship A and B sails on a
bearing of 300' for 12 km while ship B sails on a bearing of 150' for 12 km. Find

A)the distance between the 2 ships (ANs: 23.2 Km)

B)the bearing of the ship A from ship B (ANs: 315')

C)the bearing of ship B from ship A (ANs: 135')

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THe figure shows a solid which consisits of a hemishere and a right
circular cone with a common base. If its total surface are is 115丌,
find its volume.

(ANs: 576)

更新1:

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回答 (2)

momo

2008-06-25 5:08 pm

✔ 最佳答案

(A)
Let the portion be point O
angle AOB = 300 - 150 = 150
By cosine law:
AB^2 = OA^2 + OB^2 - 2(OA)(OB) cos angle AOB
AB^2 = 12^2 + 12^2 - 2(12)(12)cos 150
AB = 23.18km
(B)
By sine law:
OA / sin (angle OBA) = AB / sin (angle AOB)
12 / sin (angle OBA) = 23.18 / sin150
angle OBA = 15
Bearing from B to the port O = Bearing from port O to B + 180
= 150 + 180 = 330
Bearing from B to A = Bearing from B to port O - angle OBA
= 330 - 15 = 315
(C)
The angle OAB = 180 - angle AOB - angle OBA
= 180 - 150 - 15 = 15
Bearing from A to B = Bearing from B to A - 180
= 315 - 180 = 135

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用戶10012

2008-06-25 5:11 pm

Q.1
Let O be the port, A be ship A and B be ship B.Therefore, angle AOB = (300 - 270) + 90 + (180 - 150) = 150.
OA = OB =12, then by cosine rule, distance between A and B
= sqrt[12^2 + 12^2 -2(12)(12)cos150. = sqrt[144 + 144 -288cos150] = 23.2
Triangle AOB is an isos. triangle because OA=OB=12. Therefore, angle OBA
=(180 - 150)/2 = 15. Therefore, bearing of A from B = 360 - 30 - 15 = 315.
Bearing of B from A = 180 - (30 + 15) = 135.
Q.2
Cannot see the figure, please show it.

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