# HELP: what is value of the angle?

ABCD is a square and AD is the diameter of the circle. what is the value of $$\theta$$ ?

I think $$\theta$$ is independent of the length of the sides of the square and am getting an answer approximately 40.183 degrees. I got the answer by solving some properties of triangle equations and doing some constructions. I don't know whether it is right or wrong. My question is what is the exact angle? And can we get the angle by means of constructions only (I mean without using a calculator because in my case I needed to calculate huge surds:-\ )?

Note by Ankan Bairagi
5 months, 4 weeks ago

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Suppose the side length of the square is 1. The circle equation is:

$\Big(x - \frac{1}{2} \Big)^2 + \Big(y-1 \Big)^2 = \frac{1}{4}$

Suppose point B is the origin. Define a direction vector:

$v_x = cos \, 60^\circ \\ v_y = sin \, 60^\circ$

Substitute into the circle equation:

$\Big(\alpha \, v_x - \frac{1}{2} \Big)^2 + \Big(\alpha \, v_y-1 \Big)^2 = \frac{1}{4}$

The remaining steps are:
1) Expand to get a quadratic equation in $$\alpha$$
2) Solve for $$\alpha$$. Note that you must solve for the greater alpha value
3) $$(E_x, E_y) = (\alpha \, v_x, \alpha \, v_y)$$
4) Calculate vectors from E to B and E to C
5) Given those two vectors, use the dot product formula to calculate $$\theta$$. In other words, take the arc-cosine of a quantity equal to the dot product divided by the product of the vector magnitudes.

This yields $$\theta = 37.9235^\circ$$

- 5 months, 3 weeks ago

Taking $$B$$ as the origin is really clever

- 5 months, 3 weeks ago

Nice solution! I did a mistake in the calculation. The equations were really messy. I am also getting 37.9 degree. Thank you so much!!!!

- 5 months, 3 weeks ago

I got approximately 37.9235°, but I'm really not sure because the exact formulas get really messy.

First, I put the construction into a coordinate system with the center of the circle at the origin and 1 unit as the radiua of the circle. Every point on this circle can be written as $$(\cos \alpha, \sin \alpha)$$, where $$\alpha$$ is the central angle.
Then I considered the slope of line $$DE$$. It is given as $$60^\circ = \sqrt{3}$$, but we can also write it as $$\frac {\Delta y}{\Delta x} = \frac {\sin \alpha + 2} {\cos \alpha + 1}$$. If we set these two equal to each other, we can solve for $$\alpha$$ and get $$\alpha = 2 \arctan\left(- \frac 1 2 + \sqrt{\sqrt{3}-\frac 3 4}\right)$$.
Now we can plug this into $$E = (\cos \alpha, \sin \alpha)$$ and get $$E = \left(-\frac34+\frac12\sqrt{\sqrt{3}-\frac34}, -\frac12+\frac14\sqrt{3}+\frac12\sqrt{\sqrt{3}-\frac94}\right)$$.
Then we can calculate the slope of line $$EC$$ as $$\frac {6+\sqrt{3}+\sqrt{12\sqrt{3}-9}} {-7+2\sqrt{3}+\sqrt{4\sqrt{3}-3}}$$.
If we take the inverse tangent of this we get the angle at $$C$$ which we subtract from $$180^\circ - 60^\circ$$ to get our final answer of $$\frac{2\pi}3+\arctan\left(-\frac94-\sqrt{3}-\frac12\sqrt{\frac{81}4+12\sqrt{3}}\right) \approx 37.9235^\circ$$.

As I said, it gets really complicated and I'm not sure if I've calculated it correctly but I think the way I did it is right.

- 5 months, 3 weeks ago

I got the same answer. My solution is above

- 5 months, 3 weeks ago