# How to get the area of the triangle embedded in the unit circle when it makes an inner reflection?

The problem is as follows:

The figure represents the unit circle which has a radius of one unit. Find the value of the angle $$\alpha$$ such that the largest edge of the cherry shaded region is equal to $$\sqrt{2+\sqrt{3}}$$.

The choices given in my book are as follows:

$$\begin{array}{ll} 1.&\frac{2\pi}{3}\\ 2.&\frac{5\pi}{6}\\ 3.&\frac{11\pi}{12}\\ 4.&\frac{3\pi}{4}\\ \end{array}$$

I'm not sure exactly how to solve this problem, but I belive that the given angle alpha is referred that begins on $$A$$ and ends in the other end from the vertex of the triangle which appears to be isosceles.

I think the approach here is that to get the value of alpha I must make a trigonometric equation and use the larger edge in terms of alpha and equate this with what I was given and this may give the desired result.

The distance from the $$\textrm{x-axis}$$ to any of the vertex of the base of the triangle with those parallel to $$\textrm{y-axis}$$ is the same, hence this is an isosceles triangle and this may help or ease the solution a bit.

Since those internal angles in the circle which make up the triangle are $$\frac{\alpha}{2}$$ because it is an iscribed angle, then I'm getting:

I'm naming the end point of the angle alpha as $$P$$:

Thus $$\angle OAP=90-\alpha/2$$ and the intersection of lines where is the second little square looking from the left would be point $$R$$.

Thus:

$$\angle ROP=180-\alpha$$

Therefore:

$$OR=1\cdot \cos (180-\alpha)$$

$$AO=1$$

Thus:

$$AR=1-\cos\alpha$$

$$AR\sec \left(90-\frac{\alpha}{2}\right)= \sqrt{2+\sqrt{3}}$$

From which I believe the equation to be solved would be as follows:

$$(1-\cos\alpha)\cdot \sec \left(90-\frac{\alpha}{2}\right)= \sqrt{2+\sqrt{3}}$$

Using the necessary algebra I'm getting this into:

$$\cos\alpha=0$$

$$2\cos^2\alpha-(6+3)\cos \alpha=0$$

$$\cos\alpha=\frac{6+\sqrt{3}}{2}$$

But from looking at these expressions none of these seem to help me much into the solution.

Was my strategy wrong or what?. Can someone help me here?. I honestly don't think I'm far off from the answer, so I need a helping hand which can guide me to the right path.

First notice that the longest side of the cherry triangle (the hypotenuse) is congruent to its reflection above the $$x$$-axis: the segment connecting the initial $$(1, 0)$$ and terminal $$(\cos \alpha, \sin \alpha)$$ points of the angle $$\alpha$$ in standard position. Let's call this length $$\ell$$. Then, Pythagoras give us $$\ell^2 = (1 - \cos \alpha)^2 + (\sin \alpha)^2 = 1 - 2\cos \alpha + \cos^2 \alpha + \sin^2 \alpha = 2(1 - \cos \alpha).$$ So we need to solve the equation $$2(1 - \cos \alpha) = 2 + \sqrt{3}$$ for $$\alpha$$, i.e. $$\cos \alpha = -\frac{\sqrt{3}}{2}.$$
The diagram suggest that the terminal point is above the $$x$$-axis (can we assume this?), in which case $$0 \leq \alpha \leq \pi$$. Then, there is a unique angle $$\alpha = \arccos \biggl(\! -\frac{\sqrt{3}\,}{2} \biggr) = \frac{5\pi}{6}.$$