Question

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

Answer 1

Answer:

$\large\boxed{\sqrt3}$

Step-by-step explanation:

Look at the picture.

Use the Pythagorean theorem:

$h^2+\left(\dfrac{a}{2}\right)^2=a^2$

solve for h:

$h^2+\dfrac{a^2}{4}=a^2\qquad\text{subtract}\ \dfrac{a^2}{4}\ \text{from both sides}\\\\h^2=\dfrac{4a^2}{4}-\dfrac{a^2}{4}\\\\h^2=\dfrac{3a^2}{4}\to h=\sqrt{\dfrac{3a^2}{4}}\\\\h=\dfrac{\sqrt{3a^2}}{\sqrt4}\\\\h=\dfrac{a\sqrt3}{2}$

$tangent=\dfrac{opposite}{adjacent}$

We have:

$opposite=\dfrac{a\sqrt3}{2}\\\\adjacent=\dfrac{a}{2}$

Substitute:

$\tan60^o=\dfrac{\frac{a\sqrt3}{2}}{\frac{a}{2}}=\dfrac{a\sqrt3}{2}\cdot\dfrac{2}{a}=\sqrt3$