quick question

how is cos pi/6 = sq root 3 /2 I just can`t work it out

DL

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- September 30th 2012, 02:24 PM #1

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- September 30th 2012, 02:52 PM #2

- October 8th 2012, 04:14 AM #3

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- October 8th 2012, 05:36 AM #4

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## Re: cos pi/6

Hello, davellew69!

Consider an equilateral triangle with side length 2.

Each vertex has a 60^{o}angle.

Draw an altitude . .It bisects the angle.

The diagram looks like this.

Code:A * /|\ / | \ /30| \ 2 / |h \ 2 / | \ / | \ / 60 | \ B *-------*-------* C 1 D 1

From which we get: .

Hence, in a 30-60 right triangle, we have these sides:

. .

Now we can write all the trig values for 30^{o}and 60^{o}.