Question

Given `sin30^circ=1/2` and `tan30^circ=sqrt3/3`, how do you find `cot60^circ`?

Answer 1

`cot60°=sqrt3/3`

Explanation:

`cot60°=(cos60°)/(sin60°)`

Since

`cos60°=sin30°=1/2`

and

`sin60°=cos30°=sqrt3/2`,

you get

`cot60°=(1/cancel2)/(sqrt3/cancel2)=1/sqrt3=sqrt3/3=1/(tan60°)`