Famous Triangles
These triangles are so commonly found in geometry, trigonometry, and higher mathematics, that they deserve their own post. They form the building blocks of other more complicated structures.
30-60-90 Triangles
Consider an equilateral triangle with all side lengths of $s$. A result in geometry is that all the internal angles of a triangle add up to $180^{\circ} = \pi \ \text{rad}$. By symmetry, it must be the case that all the angles of an equilateral triangle are equal, and thus must be $60^{\circ} = \pi/3 \ \text{rad}$.
Now, draw the perpendicular bisector of one of the angles, which will cut the triangle in half. The result is a $30{-}60{-}90$ triangle.
Using the Pythagorean theorem, we can easliy compute the value of $h$.
\[\begin{align} (s/2)^2 + h^2 &= s^2 \\[10pt] \frac{1}{4}s^2 + h^2 &= s^2 \\[10pt] h^2 = \frac{3}{4} s^2 \\[10pt] h = \frac{\sqrt{3}}{2} s \end{align}\]Now, consider a $30{-}60{-}90$ triangle on the unit circle.
Thus, we’ve now learned that all of the following specific values.
\[\begin{align} &\sin 30^{\circ} = \sin \frac{\pi}{6} = \frac{1}{2} \qquad\qquad &&\sin 60^{\circ} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\\[10pt] &\cos 30^{\circ} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \qquad\qquad &&\cos 60^{\circ} = \cos \frac{\pi}{3} = \frac{1}{2}\\[10pt] &\tan 30^{\circ} = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \qquad\qquad &&\tan 60^{\circ} = \tan \frac{\pi}{3} = \sqrt{3}\\[10pt] &\sec 30^{\circ} = \sec \frac{\pi}{6} = 2 \qquad\qquad &&\sec 60^{\circ} = \sec \frac{\pi}{3} = \frac{2}{\sqrt{3}}\\[10pt] &\csc 30^{\circ} = \csc \frac{\pi}{6} = \frac{2}{\sqrt{3}} \qquad\qquad &&\csc 60^{\circ} = \csc \frac{\pi}{3} = 2\\[10pt] &\cot 30^{\circ} = \cot \frac{\pi}{6} = \sqrt{3} \qquad\qquad &&\cot 60^{\circ} = \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} \end{align}\]45-45-90 Triangles
Consider a right, isosceles triangle. Given the length of the hypotenuse, there is only one way to construct this triangle, which is given below.
Again, using the Pythagorean theorem, we can easily compute the value of $x$.
\[\begin{align} x^2 + x^2 = s^2 \\[10pt] 2x^2 = s^2 \\[10pt] x = \frac{1}{\sqrt{2}} s \end{align}\]Now, consider a $45{-}45{-}90$ triangle on the unit circle.
Thus, we’ve now learned that all of the following specific values.
\[\begin{align} &\sin 45^{\circ} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\[10pt] &\cos 45^{\circ} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\[10pt] &\tan 45^{\circ} = \tan \frac{\pi}{4} = 1 \\[10pt] &\sec 45^{\circ} = \sec \frac{\pi}{4} = \sqrt{2} \\[10pt] &\csc 45^{\circ} = \csc \frac{\pi}{4} = \sqrt{2} \\[10pt] &\cot 45^{\circ} = \cot \frac{\pi}{4} = 1 \end{align}\]