標題:
MATH Q
發問:
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. FindA)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 ... 顯示更多 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') ________________________________________________________________ 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: show your step
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最佳解答:
(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
其他解答:
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.A215E4A2B88AAE64
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