In an earlier contribution to this discussion,
wrote:

On 21 Jul, 23:13, "Roger Mills" wrote:
In an earlier contribution to this discussion,




I think it can be done (proofed) with no cos or sin involved IMHO.

Adam

Yes it can. I did it using Pythagoras and interesting chords. It
comes out as r=L/sqrt(3) - which is the same thing anyway because
cos(30)=sqrt(3)/2 --
Cheers,
Roger


My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. Why does sqrt(3)/2 = cos(30deg)? I can see that it does, I
just can't see the Pythagoras on the way to getting there......

Is it because of the defined relationship in a triangle with 30 / 60 /
90 degree angles (making the sides = 106.8, 53.4, 92.5 in this case?)

I should have done A level maths (perhaps I will one day) - its
fascintating stuff!

Matt

First off, a correction to my previous post. I've no idea what an
"interesting" chord is - I of course meant "intersecting chords"!

To answer your question. . .

Imagine an equilateral triangle with side length 2 units. Draw a
perpendicular from apex to base. You now have two 30/60/90 triangles with
the 30 degree angle at the top. The hypotenuse is 2 units long and the side
opposite the 30 degrees is 1 unit. Apply Pythagoras to that and (adjacent
side)^2 = 2^2 - 1^2 = 4 - 1 = 3
The length of the adjacent side is thus sqrt(3) units. The cosine of 30
degrees - adjacent side/hypotenuse is therefore sqrt(3)/2

QED!
--
Cheers,
Roger
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