標題:
Applications in Trigonometry
發問:
1. From a building XY, the whole circle bearings of two points A and B on the same horizontal ground are 125? and 235? respectively. The angles of elevation of the top X of the building from A and B are 33? and 47? respectively. The distance between A and B is 100m. Let XY = h m and ∠ABY = θ. ... 顯示更多 1. From a building XY, the whole circle bearings of two points A and B on the same horizontal ground are 125? and 235? respectively. The angles of elevation of the top X of the building from A and B are 33? and 47? respectively. The distance between A and B is 100m. Let XY = h m and ∠ABY = θ. (a)(i) Show that sinθ x Tan33?/ h = sin70?/ 100. (ii) What is the whole circle bearing of A from B? (Give the answers correct to the nearest degree if necessary.) http://i617.photobucket.com/albums/tt257/michaelcoco_/acbccb87.jpg
最佳解答:
1. From a building XY, the whole circle bearings of two points A and B on the same horizontal ground are 125? and 235? respectively. The angles of elevation of the top X of the building from A and B are 33? and 47? respectively. The distance between A and B is 100m. Let XY = h m and ∠ABY = θ. (a) (i) Show that sinθ x Tan33?/ h = sin70?/ 100. In △AXY, tan33?= h / AY AY = h / tan33? In △ABX, AY / sin∠ABY = AB / sin∠AYB ( h / tan33?) / sinθ = 100 / sin ( 235?- 125?) ( h / ( tan33?x sinθ ) = 100 / sin110? sinθ x tan33?/ h= sin ( 180?- 110?) / 100 sinθ x tan33?/ h= sin70?/ 100 (ii) What is the whole circle bearing of A from B? In △BXY, tan47?= h / BY BY = h / tan47? In △ABX, AB2 = AY2 + BY2 – 2 AY x BY cos∠AYB 1002 = ( h / tan33?)2 + (h / tan47?)2 – 2 (h / tan47?) ( h / tan33?) cos110? 1002 = ( 2.3712 + 0.8696 + 0.9822 ) h2 h = 48.66 From (1) : sinθ x tan33?/ h= sin70?/ 100 sinθ = h sin70?/ ( 100 tan33?) = 48.66 sin70?/ ( 100 tan33?) = 0.7041 θ = 44.76? The whole circle bearing of A from B is : 235?- 180?+ 44.76? = 99.76?= 100? (correct to the nearest degree)
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