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 $\sqrt{2+\sqrt{3}}$.

Sketch of the problem

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 $R$.


$\angle ROP=180-\alpha$


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




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

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:


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


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 $$ \alpha = \arccos \biggl(\! -\frac{\sqrt{3}\,}{2} \biggr) = \frac{5\pi}{6}. $$


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