Question

Given `sin60^circ=sqrt3/2` and `cos60^circ=1/2`, how do you find `cot60^circ`?

Answer 1

`=0.577` or `1/sqrt3`

Explanation:

`cot60^@=cos60^@/sin60^@`

`=(1/2)/(sqrt3/2)`

`=1/sqrt3`

`=sqrt3/3`

`=0.577`