Use a special right triangle to write tan 60° as a fraction.

Answer:[tex]\large\boxed{\sqrt3}[/tex]Step-by-step explanation:Look at the picture.Use the Pythagorean theorem:[tex]h^2+\left(\dfrac{a}{2}\right)^2=a^2[/tex]solve for h:[tex]h^2+\dfrac{a^2}{4}=a^2\qquad\text{subtract}\ \dfrac{a^2}{4}\ \text{from both sides}\\\\h^2=\dfrac{4a^2}{4}-\dfrac{a^2}{4}\\\\h^2=\dfrac{3a^2}{4}\to h=\sqrt{\dfrac{3a^2}{4}}\\\\h=\dfrac{\sqrt{3a^2}}{\sqrt4}\\\\h=\dfrac{a\sqrt3}{2}[/tex][tex]tangent=\dfrac{opposite}{adjacent}[/tex]We have:[tex]opposite=\dfrac{a\sqrt3}{2}\\\\adjacent=\dfrac{a}{2}[/tex]Substitute:[tex]\tan60^o=\dfrac{\frac{a\sqrt3}{2}}{\frac{a}{2}}=\dfrac{a\sqrt3}{2}\cdot\dfrac{2}{a}=\sqrt3[/tex]