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 ${tan 60}^{\circ}$ $tan60^circ$ ?

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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 ${tan 60}^{\circ}$ $tan60^circ$ ?

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Answer 1 · 2018-08-10T12:34:35

${tan 60}^{\circ} = \sqrt{3}$ $tan60^@=sqrt3$

Explanation:

$using the trigonometric identity$ $"using the "color(blue)"trigonometric identity"$

$∙ x tan θ = \frac{sin θ}{cos θ}$ $•color(white)(x)tantheta=sintheta/costheta$

$thus$ $"thus"$

${tan 60}^{\circ} = \frac{{sin 60}^{\circ}}{{cos 60}^{\circ}}$ $tan60^@=sin60^@/cos60^@$

$color(white)(xxxxx)=(sqrt3/2)/(1/2)=sqrt3/cancel(2)xxcancel(2)/1=sqrt3$

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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 ${tan 60}^{\circ}$ $tan60^circ$ ?

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Given sin60^circ=sqrt3/2sin60∘=√32sin60^circ=sqrt3/2 and cos60^circ=1/2cos60∘=12cos60^circ=1/2, how do you find tan60^circtan60∘tan60^circ?

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Explanation:

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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 ${tan 60}^{\circ}$ $tan60^circ$ ?