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maths唔識做..
發問:
3.) Given that tanθ=3/4 , evaluate (3sinθ - 4cosθ)/(3sinθ+ 4cosθ)4.) Two ships A and B leave a port at the same time. Ship A 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 ship A & ship B,(b)the... 顯示更多 3.) Given that tanθ=3/4 , evaluate (3sinθ - 4cosθ)/(3sinθ+ 4cosθ) 4.) Two ships A and B leave a port at the same time. Ship A 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 ship A & ship B, (b)the bearing of ship A from ship B, (c)the bearing of ship B from ship A. thx!
最佳解答:
3.) Given that tanθ = 3/4 , evaluate (3sinθ - 4cosθ)/(3sinθ+ 4cosθ) ( 3 sinθ – 4 cosθ ) / ( 3 sinθ + 4 cosθ ) = [ ( 3 sinθ – 4 cosθ ) / cosθ ] / [ ( 3 sinθ + 4 cosθ ) / cosθ ] = ( 3 sinθ / cosθ – 4 ) / ( 3 sinθ / cosθ + 4 ) = ( 3 tanθ – 4 ) / ( 3 tanθ + 4 ) = ( 3 x 3 / 4 – 4 ) / ( 3 x 3 / 4 + 4 ) = ( 9 / 4 – 4 ) / ( 9 / 4 + 4 ) = [ ( 9 – 16 ) / 4 ] / [ ( 9 + 16 ) / 4 ] = ( 9 – 16 ) / ( 9 + 16 ) = -7 / 25 4.) Two ships A and B leave a port at the same time. Ship A 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 ship A & ship B, Two ships A and B leave a port O at the same time. ∠AOB = 300?– 150?= 150? OA = OB = 12 km △AOB is an isos. triangle. Draw OC perpendicular AB and meet AB at C. In rt∠△ACO, ∠AOC = (1/2) ∠AOB = 150?/ 2 = 75? sin∠AOC = AC / OA sin75?= AC / 12 AC = 12 sin75?= 11.591 AB = 2 AC = 44.785 x 2 = 23.182 The distance between ship A & ship B is 23.18 km (b) the bearing of ship A from ship B, ∠OBA = ( 180?– 150?) / 2 = 30?= 15? The bearing of ship A from ship B is : 360?– 30?– 15?= 315? (c) the bearing of ship B from ship A. The bearing of ship B from ship A is : 90?+ 30?+ 15?= 135?
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