Question

Use a special right triangle to find the tangent of 30∘
Tan 30*= ?

Answer 1

The value of tan of 30 degrees is given by:

$tan(30^\circ) = \dfrac{1}{\sqrt{3}}$

What is a right angled triangle?

A right angled triangle is a triangle containing at least one angle with 90 degrees measurement.

We will use equilateral triangle( since its all angles are of 60 degrees) and split it into two composing right angled triangle using  a perpendicular from A to line BC.

Since we have to find the value at 30 degrees, thus we will bisect the 60 degree angle in two parts as shown in the below given diagram.

Using angle bisector theorem

" In a triangle ABC, if a line passes through vertex A and touches BC at point D, then    $\dfrac{BD}{CD} = \dfrac{AB}{AC}$ "

Dropping a line from A to side AD at point E such that the angle A is cut into two equal parts each of 30 degrees. Then we have by angle bisector theorem for triangle ABD:

$\dfrac{ED}{AE} = \dfrac{BD}{AB} = \dfrac{\dfrac{x}{2}}{x} = \dfrac{1}{2}\\\\AE = 2ED$

$H^2 = P^2 + B^2\\\\AB^2 = BD^2 + AD^2\\\\AD^2 = x^2 - (\dfrac{x}{2})^2 = \dfrac{3x^2}{4}\\\\AD = \dfrac{x}{2}\sqrt{3}$

Using Pythagoras Theorem

Since AD = AE + ED and since AE = 2ED, thus we have:

$\dfrac{x}{2}\sqrt{3} = AE + ED = 2ED + ED = 3ED\\\\ED = \dfrac{x}{6}\sqrt{3}\\\\AE = 2ED = \dfrac{x}{3}\sqrt{3}$

In triangle BDE, we have $\angle D = 90^\circ$, and $\angle EBD = \dfrac{60}{2} = 30^\circ$, thus we have:

$tan(30^\circ) = \dfrac{ED}{BD} = \dfrac{\dfrac{x}{6}\sqrt{3}}{\dfrac{x}{2}} = \dfrac{1}{3}\sqrt{3} = \dfrac{1}{\sqrt{3}}$

Thus, the value of tan of 30 degrees is given by:

$tan(30^\circ) = \dfrac{1}{\sqrt{3}}$

Learn more about trigonometric ratios here:

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