## 45-45-90 You could use [[Pythagorean Theorem]] Duplicate a reflection of a 45-45-90 triangle over one of its legs ![[45-45-90-triangle.svg|-noinv-dmo]] The two traigngles are part of a larger similar triangle. The proportion of the scale factor is: $\frac{1}{2x}=\frac{x}{1} \implies \frac{1}{2}=x^2 \implies x=\frac{\sqrt{2}}{2}$ ## 30-60-90 ![[30-60-90-triangle.svg|-noinv-dmo]] ## 15-75-90 Start with a [[#45-45-90]] and cut one of the $45\degree$ angles into $30\degree$ and $15\degree$. The segment that makes this cut splits the opposite side into side lengths of special right triangles [[#45-45-90]] and [[#30-60-90]]. This can be used to find the side lengths of the inner [[#15-75-90]] triangle, and then it can be scaled so that the hypotenuse is 1. ![[15-75-90-triangle.svg|-noinv-dmo]] ## 18-72-90 Start with a 36-72-72 isosceles triangle. Let the long side be 1, and bisect one of the $72\degree$ angles to create a similar triangle to the original. ![[36-72-72-isos.svg|-dmo-noinv]] The similarity of the triangles provides a proportion that can be solved with quadratic formula: $\begin{aligned} \frac{x}{1}&=\frac{1-x}{x}\\ \\ x^2&=1-x\\ \\ x^2+x-1&=0\\ \\ \text{(only the positive) }x&=\frac{-1+\sqrt{5}}{2} \end{aligned}$ Now the isosceles triangle is cut in half by its height to create the 18-72-90 ![[36-72-72-into-18-72-90.svg|-noinv-dmo]] And using [[Pythagorean Theorem]] $\begin{aligned} h&=\sqrt{1^2+\left(\frac{-1-\sqrt{5}}{4}\right)^2}\\ h&=\sqrt{1-\frac{6-2\sqrt5}{16}}\\ h&=\sqrt{\frac{10+2\sqrt5}{16}} \\ h&=\frac{\sqrt{10+2\sqrt5}}{4} \end{aligned}$ ## Summary ![[4-special-right-triangles.svg|-dmo-noinv]]