A drawbridge AB of length 20m is streched across a river horizontally . It is now rotated upwards about A through an abgle of 40 degree to a new position AC as shown in the figure. Find the distance between
a. C and D
b. B and C
http://flypicture.com/display/MjE0NTQ=
F2 Maths ( 15marks)
2006-12-10 4:50 am
回答 (3)
2006-12-10 5:19 am
✔ 最佳答案
sin 角 CAD= CD / 20 sin 40∘= CD / 20
CD = 20 sin 40∘
=12.85575219m
因為係同一條bridge轉左上去，所以 AB = AC ，三角形 ABC係一個等邊三角形。
角 ABC = 角 ACB = (180∘- 40∘) / 2 = 70∘
sin 角 ABC= CD / BC
sin 70∘= CD / BC
sin 70∘= ( 20 sin 40∘) / BC
BC = ( 20 sin 40∘) / sin 70∘
= 13.68080573m
2006-12-10 5:22 am
(a)
AB = AC = 20m (because AC is just the new position of AB after rotation)
sin 40 = CD/AC
sin 40 = CD/20
CD= 20sin40
= 12.8558m
(b)
because AB = AC = 20m,
therefore triangle ABC is an isosceles triangle.
angle DBC = (180-angle CAD)/2 (base angles, isos. triangle)
angle DBC = (180-40)/2
= 70
from (a), CD = 12.8558m
sin (angle DBC) = CD/CB
sin 70 = 12.8558/CB
CB = 12.8558/sin70
= 13.6809m
AB = AC = 20m (because AC is just the new position of AB after rotation)
sin 40 = CD/AC
sin 40 = CD/20
CD= 20sin40
= 12.8558m
(b)
because AB = AC = 20m,
therefore triangle ABC is an isosceles triangle.
angle DBC = (180-angle CAD)/2 (base angles, isos. triangle)
angle DBC = (180-40)/2
= 70
from (a), CD = 12.8558m
sin (angle DBC) = CD/CB
sin 70 = 12.8558/CB
CB = 12.8558/sin70
= 13.6809m
2006-12-10 5:03 am
a) AC=AB=20
CD=AB x sin40
=20sin40
b)angleDBC=(180-40)/2
BC xsinDBC=CD
BC=CD/sinDBC
CD=AB x sin40
=20sin40
b)angleDBC=(180-40)/2
BC xsinDBC=CD
BC=CD/sinDBC
收錄日期: 2021-04-13 00:33:02
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