Summation of infinite cos series and determining theta

Question

Question: In the figure,

$A_0A_1,A_2A_3,A_4A_3...$ are all perpendicular to $L_1$

$A_1A_2,A_3A_4,A_5A_6...$ are all perpendicular to $L_2$

If $A_0A_1=1$

And $A_0A_1+A_1A_2+A_2A_3+A_3A_4......\infty =2(2+ \sqrt3)$

Find $\theta$

I figured that all the triangles formed are similar and hence the $\cos θ$ is constant.

Now for finding the sum, $A_0A_1=A_1A_2=... =\ cos θ$

$S = \cos θ+\cos θ+\cos θ+....\mathrm{(infinite\ times)} = 2(2+ √3)$

how can I determine $\theta$?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Commented Dec 31, 2023 at 15:08
an infinite sum of the same constant cannot converge to a finite value other than $0$ — Ryan Tay, Commented Dec 31, 2023 at 15:26

Dheeraj Gujrathi · Accepted Answer · 2023-12-31 17:15:55Z

Your mistake is critical, First of all, $A_0A_1=1\neq \cos \theta$ , If $\theta\neq0$, Which should be necessary, So that triangle is formed

Now,

$$A_1A_2=\cos \theta$$

$$A_2A_3=\cos^2\theta$$

$$A_3A_4=\cos^3\theta$$

$$A_4A_5=\cos^4\theta$$

$$.$$$$.$$$$.$$

$$.$$

hence the sum $$S=1+\cos\theta+\cos^2\theta+\cos^3\theta....$$

which Is a $G.P$

Hence

$$S=\frac{1}{1-\cos \theta}$$

which leads to

$$\frac{1}{1-\cos \theta}=2(2+\sqrt3)$$

Solving for $\cos \theta$ yields

$$\cos \theta=\frac{\sqrt3}{2}$$

The only possible angle for a triangle gives $$\theta =\frac{\pi}{6}$$Where $\theta$ is in radians

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