Hello, I want to find the angle accurately. Please help me out.

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:-\ )?

Thanks in advance.

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## Comments

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TopNewestSuppose 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 value3)\((E_x, E_y) = (\alpha \, v_x, \alpha \, v_y) \)4)Calculate vectors from E to B and E to C5)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 \(\theta = 37.9235^\circ \)

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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!!!!

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Taking \( B \) as the origin is really clever

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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.

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I got the same answer. My solution is above

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