Making a right-angled 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.

Making another right-angled triangle out of CD and the corner, we now know the side length and we want to know the hypotenuse. cosθ=a/h, so cos 60=0.1503.../h, so h=0.1503.../cos 60=0.30m to the nearest cm, which is length CD.

The outer frame has two lengths and two widths on it, so in total it is 10+10+0.8+0.8=21.6m long and one length AB is 0.8/sin 60=0.9238...m, so 21 of them are 19.40m to the nearest cm.

1=cos^2 x+sin^2 x
then, group all the sin terms...

2 sinx cosx +cos^2x=cos^2x +sin^2 x
2 sinx cosx=sin^2 x
sin^2 x-2 sinx cos x=0
sin x(sin x-2 cos x)=0
sin x=0 or sin x=2 cos x
x=0,etc..
or tan x=2
x=...

You are right!
When x=0,
2sinx.cosx + cos^2x = 1
But how do we get a solution x=0?
2sinx.cosx + cos^2x = 1
sin2x + cos^2x = 1,
sin2x = 1-cos^2x
sin2x = sin^2x
this would be true only when x=0
But is there any (other)mathematical way it can be shown that x=0 is a solution?

thanks ajp3. But I got another question on trigo...

Given cosec A + cot A = 3

and thus evalute cosec A -cot A

and find cos A

Last edited by ahgua (2005-08-26 20:52:31)

Prove that (sec^2 x + 2 tan x)/ (1 - tan^2 x) = (cos x + sin x)/(cos x - sin x)
Hence solve the equation (sin x +cos x)/(sin x - cos x) = tan x for 0 < x < 2pi.