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04 Apr 2010, 12:46
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hi can u explain me the soln thks
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Updated on: 04 Apr 2010, 19:48
Originally posted by achiever01 on 04 Apr 2010, 19:22.
Last edited by achiever01 on 04 Apr 2010, 19:48, edited 1 time in total.
Last edited by achiever01 on 04 Apr 2010, 19:48, edited 1 time in total.
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Please see the attached doc.. it will be much clearer...though i am not too good in drawing the diagrams in msword
BC = 54(length of the drawbridge)
BD=CE=28( as its a 2 m overlap, it means, 52/2+2=28)
Extend BD and CE so that it meets in A.
ABC is an equilateral triangle as <ABC=<ACB=60 =><BAC=60
hence AB=BC=54
=>DA=EA = 54-28 = 26
Triangle ADE is an equilateral triangle as <ADE=AED=60
hence clearance width = DE = DA = 26.
HTH.
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triangle.JPG [ 16.1 KiB | Viewed 1761 times ]
04 Apr 2010, 19:36
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Attachment:
Untitled.jpg [ 8.05 KiB | Viewed 1779 times ]
The channel width ,clearance width and the raisable sections form a trapezoid with channel width and clearance width as parallel bases and the raisable sections as the legs.
I just redrew the trapezoid above with vertices as ABCD where CD is the channel width (54m) and BC and AD are the raisable sections. We have to find the clearance width which is AB
The channel width is given as 54 m and since the raisable sections are equal in size and they overlap each other by 2 m when they are closed
sum of lengths of raisable sections = 54 + 2. Each raisable section length is equal to 28m . So BC=AD =28m
If we draw altitude from vertices A and B to CD ABFE forms a rectangle since AB and CD are parallel.
Since ABFE is a rectangle AB=FE and AF=BE. So if we find FE that will be equal to our clearance width.
Also AFD and BEC form right triangles with right angle at F and E respectively and <ADF = <BCE = 60 degrees as per the problem. So AFD and BEC are 30-60-90 degree triangles.
In a 30-60-90 degree triangles the sides will be in ratio of 1:\sqrt{3}:2
1 corresponding to the side opposite 30 degrees , \sqrt{3} corresponding to the side opposite 60 degrees and 2 corresponding to the side opposite 90 degrees.
We already know AD (which the length of raisable section) so DF = 28/2 =14m and AF= 14\sqrt{3} m . Similarly EC=14m and BE=14\sqrt{3} m
FE= CD - (DF + EC)
= 54 -(14 +14)
= 26m
So the clearance width is 26m
