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

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Answer 1 · 2017-05-24T14:39:22

$cot60°=sqrt3/3$

$cot60°=(cos60°)/(sin60°)$

Since

$cos60°=sin30°=1/2$

and

$sin60°=cos30°=sqrt3/2$ ,

you get

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

Given sin30^circ=1/2sin30^circ=1/2 and tan30^circ=sqrt3/3tan30^circ=sqrt3/3, how do you find cot60^circcot60^circ?