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°)#