# How do you find trigonometric ratios of 30, 45, and 60 degrees?

Jun 25, 2015

The trigonometric ratios for ${30}^{o}$, ${45}^{o}$, and ${60}^{o}$ are based on some standard triangles. sin, cos, and tan (and their reciprocals) are the ratios of the sides of these triangles.

#### Explanation:

Both ${30}^{o}$ and ${60}^{o}$ are based on an equilateral triangle with sides of length 2 and with one of the angles bisected.

The ${45}^{o}$ angle is based on an isosceles triangle with the equal sides having a length of 1.

For all triangles the Pythagorean Theorem is used to compute the "missing" side length.

If remember that
$\textcolor{w h i t e}{\text{XXXX}}$sin $= \text{opposite"/"hypotenuse}$

$\textcolor{w h i t e}{\text{XXXX}}$cos $= \text{adjacent"/"hypotenuse}$

$\textcolor{w h i t e}{\text{XXXX}}$tan $= \text{opposite"/"adjacent}$

and their reciprocals.

Then, for example:
$\textcolor{w h i t e}{\text{XXXX}}$$\sin \left({60}^{o}\right) = \frac{\sqrt{3}}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$\sin \left({30}^{o}\right) = \frac{1}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$\sin \left({45}^{o}\right) = \frac{1}{\sqrt{2}}$

$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left({60}^{o}\right) = \frac{1}{2}$

etc.

Jun 4, 2018

The other answer is fine. I'll just point out that once the student has internalized these Two Tired Triangles of Trig (30/60/90 and 45/45/90) they're done. If you review the trig questions here you'll find the vast majority use just these triangles.