The Triangle Problem

Discover a simple yet engaging trigonometry problem that involves an isosceles triangle within a square. Your goal is to determine the length of the triangle’s base given the angle of the isosceles triangle and the side length of the square. Sharpen your problem-solving skills and enjoy exploring the relationship between angles and side lengths in this intriguing challenge. Have fun uncovering the solution!

Statement

Find a function $f$ of $a$ and $\theta$ with domain

$$f:{\mathbb{R}}^{+}\times (0,\frac{\pi }{2}]\to {\mathbb{R}}^{+}$$

such that $f(a,\theta )=x$.

For example, consider $a=1$ and $\theta =\frac{\pi }{2}$, then the value of $x$ is

$$x=f(1,\frac{\pi }{2})=\sqrt{1^2+1^2}=\sqrt{2}$$

Solution

Defining $x$ in terms of $b$

First, let’s define a new variable $b$ equal to the length of the two remaining sides of the triangle

We can now write $x$ in terms of $b$ and $\theta$

$$\begin{array}{rl}\sin (\frac{\theta }{2})& =\frac{x/2}{b}\\ x& =2b\sin (\frac{\theta }{2})\end{array}$$

Finding $b$ value

Now we have to find the value of $b$

$$\begin{array}{rl}\cos (\beta )& =\frac{a}{b}\\ b& =\frac{a}{\cos (\beta )}\end{array}$$

To calculate $\cos (\beta )$, keep in mind that $2\beta +\theta =\frac{\pi }{2}$

$$\begin{array}{rl}2\beta +\theta & =\frac{\pi }{2}\\ 2\beta & =\frac{\pi }{2}-\theta \\ \beta & =\frac{\frac{\pi }{2}-\theta }{2}\end{array}$$

Then

$$\begin{array}{rl}\cos (\beta )& =\cos (\frac{\frac{\pi }{2}-\theta }{2})\\ & =\cos (\frac{\theta -\frac{\pi }{2}}{2})\\ & =\cos (\frac{\theta }{2}-\frac{\pi }{4})\\ & =\cos (\frac{\theta }{2})\cos (\frac{\pi }{4})+\sin (\frac{\theta }{2})\sin (\frac{\pi }{4})\\ & =\frac{1}{\sqrt{2}}[\cos (\frac{\theta }{2})+\sin (\frac{\theta }{2})]\end{array}$$

and

$$b=\frac{a\sqrt{2}}{\cos (\frac{\theta }{2})+\sin (\frac{\theta }{2})}$$

Finding $x$

$$\begin{array}{rl}x& =2b\sin (\frac{\theta }{2})\\ & =2\sqrt{2}a\frac{\sin (\frac{\theta }{2})}{\sin (\frac{\theta }{2})+\cos (\frac{\theta }{2})}\\ & =\sqrt{8}a(\frac{\sin (\frac{\theta }{2})+\cos (\frac{\theta }{2})}{\sin (\frac{\theta }{2})}{)}^{-1}\\ & =\sqrt{8}a(1+\frac{\cos (\frac{\theta }{2})}{\sin (\frac{\theta }{2})}{)}^{-1}\\ & =\sqrt{8}a\frac{1}{1+\cot (\frac{\theta }{2})}\\ \therefore x& =\frac{\sqrt{8}a}{1+\cot (\frac{\theta }{2})}\end{array}$$