Given ${sin 60}^{\circ} = \frac{\sqrt{3}}{2}$ $sin60^circ=sqrt3/2$ and ${cos 60}^{\circ} = \frac{1}{2}$ $cos60^circ=1/2$ , how do you find ${cot 60}^{\circ}$ $cot60^circ$ ?

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Answer 1 · 2017-01-11T06:04:43

$= 0.577$ $=0.577$ or $\frac{1}{\sqrt{3}}$ $1/sqrt3$

${cot 60}^{\circ} = \frac{{cos 60}^{\circ}}{{sin 60}^{\circ}}$ $cot60^@=cos60^@/sin60^@$

$= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$ $=(1/2)/(sqrt3/2)$

$= \frac{1}{\sqrt{3}}$ $=1/sqrt3$

$= \frac{\sqrt{3}}{3}$ $=sqrt3/3$

$= 0.577$ $=0.577$

Given sin60∘=√32sin60^circ=sqrt3/2 and cos60∘=12cos60^circ=1/2, how do you find cot60∘cot60^circ?