695. Find the slope of the line that forms an angle of $60^{\circ}$ with the horizontal.
$\triangle$ In a right triangle with an angle of $60^{\circ}$, the smaller leg is equal to half the hypotenuse (complete to an equilateral triangle, problem 167). By taking the smaller leg to be 1, we can find the larger leg using the Pythagorean theorem: it is equal to $\sqrt{2^{2}-1^{2}}=\sqrt{3}$. Thus, the slope of the line going at an angle of $60^{\circ}$ is $\sqrt{3} . \triangleleft$