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You are not logged in. #1 20050823 21:44:53
Hep with Trig. please :)The structure below measures 10m in length by 0.8m in width. In total, there are 21 diagonal pieces exactly like AB and two extra pieces CD. Pieces CD are parallel to pieces AB. Pieces AB are inclined at 60 degrees from the horizontal. #2 20050823 22:24:41
Re: Hep with Trig. please :)Making a rightangled triangle out of AB, the line going down from B and the line going right from A, we have one angle and one length. The 'opposite' side is 0.8m, the angle is 60° and we want to find the 'adjacent' side. tanθ=o/a, so tan 60=0.8/a, so a=0.8/tan 60=0.461... There are 21 of these, so 9.6995... of the 10m is being taken up by them. That leaves 0.3005... for CD. There are 2 of CD, so they will get 0.1503... each. Why did the vector cross the road? It wanted to be normal. #5 20050824 20:59:24
Re: Hep with Trig. please :)
Are you sure we can divide sin x by cos x? #6 20050824 22:16:35
Re: Hep with Trig. please :)You are right! Character is who you are when no one is looking. #7 20050826 11:31:05
Re: Hep with Trig. please :)ahgua, do not worry about dividing by cos x in the (correct) sol'n given by wcy; sin x/cos x is defined as tan x, so we do not need to worry about division by zero, the dfn of tan x already forbids it! #9 20050826 22:09:31
Re: Hep with Trig. please :)cosecA + cotA = 3 Character is who you are when no one is looking. 