quick question

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

DL

Results 1 to 5 of 5

- Sep 30th 2012, 03:24 PM #1

- Joined
- Nov 2011
- From
- Chelmsford, Essex, UK
- Posts
- 67

- Sep 30th 2012, 03:52 PM #2

- Oct 8th 2012, 05:14 AM #3

- Joined
- Oct 2012
- From
- Germany
- Posts
- 27
- Thanks
- 4

- Oct 8th 2012, 06:36 AM #4

- Oct 8th 2012, 07:18 AM #5

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 847

## 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}.