# Special Right Triangles

       

The steps to obtain the six trigonometric ratios of the special angles $30^{\circ}, 45^{\circ}$ and $60^{\circ}$ using special right triangles are presented.
The values of the six trigonometric of the special angles are given in a table at the bottom of the page.

## Isosceles Right Triangle or 45-45-90 Triangle

It is a right triangle with equal sides and angles equal to $45^{\circ}$ as shown in figure 1 below. Figure 1

This triangle helps in obtaining trigonometric ratios of a $45^{\circ}$ angle. We first use the
Pythagorean theorem to find the hypotenus $h$.

$\qquad h^2 = a^2 + a^2$
Simplify and solve foe $h$
$\qquad h^2 = 2 a^2$
$\qquad h = a \sqrt 2$
Let us now use the above triangle, apply the six trigonometric ratios to find all six trigonometric ratios of an angle of $45^{\circ}$.

$\qquad \sin 45^{\circ} = \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = \dfrac{a}{h} = \dfrac{a}{a \sqrt 2} = \dfrac{1}{ \sqrt 2} = \dfrac{\sqrt 2}{2}$

$\qquad \cos 45^{\circ} = \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = \dfrac{a}{h} = \dfrac{a}{a \sqrt 2} = \dfrac{1}{ \sqrt 2} = \dfrac{\sqrt 2}{2}$

$\qquad \tan 45^{\circ} = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} = \dfrac{a}{a} = 1$

$\qquad \csc 45^{\circ} = \dfrac{1}{\sin 45^{\circ}} = \dfrac{2}{\sqrt 2} =\sqrt 2$

$\qquad \sec 45^{\circ} = \dfrac{1}{\cos 45^{\circ}} = \dfrac{2}{\sqrt 2} = \sqrt 2$

$\qquad \cot 45^{\circ} = \dfrac{1}{\tan 45^{\circ}} = \dfrac{1}{1} = 1$

## 30-60-90 Right Triangle

We start with an equilateral triangle with side $a$ as shown in figure 2 below. Then draw a perpendicular from one of the vertices of the triangle to the opposite base. This perpendicular bisects the angle into two equal angles of $30^{\circ}$ and the opposite side into two equal segments of length $\dfrac{a}{2}$ as shown in figure 3 below. Figure 2 Figure 3

This special right triangle 30-60-90 in figure 3 helps us find the six trigonometric ratios of angles $30^{\circ}$ and $60^{\circ}$.

We first use the Pythagorean theorem to find the side $h$.

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

Solve for $h^2$.
$\qquad h^2 = a^2 - \dfrac{a^2}{4} = 3 \dfrac{a^2}{4}$
Take the square root of both sides to solve for $h$.
$\qquad h = \dfrac{a \sqrt 3}{2}$

We now use the above triangle to find all six trigonometric ratios of $30^{\circ}$.

$\qquad \sin 30^{\circ} = \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = \dfrac{\dfrac{a}{2}}{a} = \dfrac{1}{2}$

$\qquad \cos 30^{\circ} = \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = \dfrac{h}{a} = \dfrac{\dfrac{a \sqrt 3}{2}}{a} = \dfrac{\sqrt 3}{2}$

$\qquad \tan 30^{\circ} = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} = \dfrac{\dfrac{a}{2}}{h} = \dfrac{\dfrac{a}{2}}{\dfrac{a \sqrt 3}{2}} = \dfrac{1}{\sqrt 3}$

$\qquad \csc 30^{\circ} = \dfrac{1}{\sin 30^{\circ}} = \dfrac{1}{1/2} = 2$

$\qquad \sec 30^{\circ} = \dfrac{1}{\cos 30^{\circ}} = \dfrac{2}{\sqrt 3}$

$\qquad \cot 30^{\circ} = \dfrac{ 1 }{\tan 30^{\circ} } = \sqrt 3$

We now use the same triangle in figure 3 to find all six trigonometric ratios of $60^{\circ}$.
$\qquad \sin 60^{\circ} = \dfrac{\text{Opposite Side}}{\text{Hypotenuse}} = \dfrac{h}{a} = \dfrac{\dfrac{a \sqrt 3}{2}}{a} = \dfrac{\sqrt 3}{2}$

$\qquad \cos 60^{\circ} = \dfrac{\text{Adjacent Side}}{\text{Hypotenuse}} = \dfrac{a/2}{a} = \dfrac{1}{2}$

$\qquad \tan 60^{\circ} = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}} = \dfrac{h}{a/2} = \dfrac{\dfrac{a \sqrt 3}{2}}{a/2} = \sqrt 3$

$\qquad \csc 60^{\circ} = \dfrac{1}{\sin 60^{\circ}} = \dfrac{1}{\dfrac{\sqrt 3}{2}} = \dfrac{2}{\sqrt 3}$

$\qquad \sec 60^{\circ} = \dfrac{1}{\cos 60^{\circ}} = \dfrac{1}{1/2} = 2$

$\qquad \cot 60^{\circ} = \dfrac{ 1 }{\tan 60^{\circ} } = \dfrac{1}{\sqrt 3}$

## Table of the Six Trigonomeotric Functions for Special Angles

Here we group all values of the six trigonometric function in a table.
NOTE that the letter $U$ used in the table mean undefined.

 $\color{red}{\theta \;\;\; \text{( in Degrees )} }$ $\color{red}{ 0^{\circ} }$ $\color{red}{ 30^{\circ} }$ $\color{red}{ 45^{\circ} }$ $\color{red}{ 60^{\circ} }$ $\color{red}{ 90^{\circ} }$ $\color{red}{\theta \;\;\; \text{( in Radians )} }$ $\color{red}{ 0 }$ $\color{red}{ \dfrac{\pi}{6} }$ $\color{red}{ \dfrac{\pi}{4} }$ $\color{red}{ \dfrac{\pi}{3} }$ $\color{red}{ \dfrac{\pi}{2} }$ $\color{red}{\sin \theta}$ $0$ $\dfrac{1}{2}$ $\dfrac{\sqrt 2}{2}$ $\dfrac{\sqrt 3}{2}$ $1$ $\color{red}{\cos \theta}$ $1$ $\dfrac{\sqrt 3}{2}$ $\dfrac{\sqrt 2}{2}$ $\dfrac{1}{2}$ $0$ $\color{red}{ \tan \theta }$ $0$ $\dfrac{1}{\sqrt 3}$ $1$ $\sqrt 3$ $\text{U}$ $\color{red}{\csc \theta}$ $\text{U}$ $2$ $\sqrt 2$ $\dfrac{2}{\sqrt 3}$ $1$ $\color{red}{\sec \theta}$ $1$ $\dfrac{2}{\sqrt 3}$ $\sqrt 2$ $2$ $\text{U}$ $\color{red}{ \cot \theta }$ $\text{U}$ $\sqrt 3$ $1$ $\dfrac{1}{\sqrt 3}$ $0$