how is cos pi/6 = sq root 3 /2 I just can`t work it out
Consider an equilateral triangle with side length 2.
Each vertex has a 60o angle.
Draw an altitude . .It bisects the angle.
The diagram looks like this.
Pythagorus says: .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 30o and 60o.