Jse a special right triangle to write \( \tan 60^{\circ} \) as a fraction. \( \frac{\sqrt{3}}{2} \) \( \frac{\sqrt{3}}{\frac{\sqrt{3}}{3}} \) 1

In a 30-60-90 triangle, the sides are in a special ratio: the lengths are \(1\) (opposite the \(30^{\circ}\) angle), \(\sqrt{3}\) (opposite the \(60^{\circ}\) angle), and \(2\) (the hypotenuse). To find \(\tan 60^{\circ}\), we take the ratio of the opposite side to the adjacent side, which is \(\frac{\sqrt{3}}{1} = \sqrt{3}\). Moreover, here's a fun fact: the tangent function can be visualized on a unit circle! At an angle of \(60^{\circ}\), the coordinates of the point on the unit circle are \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\). The tangent, which is the y-coordinate divided by the x-coordinate, gives you \(\tan 60^{\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\). Who knew geometry could be so enlightening?