HELP: what is value of the angle?

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 θ \theta ?

I think θ \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.

Note by Ankan Bairagi
1 year ago

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1 vote

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

(x12)2+(y1)2=14 \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:

vx=cos60vy=sin60v_x = cos \, 60^\circ \\ v_y = sin \, 60^\circ

Substitute into the circle equation:

(αvx12)2+(αvy1)2=14 \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 α\alpha
2) Solve for α\alpha. Note that you must solve for the greater alpha value
3) (Ex,Ey)=(αvx,αvy)(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 θ\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 θ=37.9235\theta = 37.9235^\circ

Steven Chase - 1 year ago

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Taking B B as the origin is really clever

Henry U - 1 year ago

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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!!!!

Ankan Bairagi - 1 year ago

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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 (cosα,sinα) (\cos \alpha, \sin \alpha) , where α \alpha is the central angle.
Then I considered the slope of line DE DE . It is given as 60=3 60^\circ = \sqrt{3} , but we can also write it as ΔyΔx=sinα+2cosα+1 \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 α \alpha and get α=2arctan(12+334) \alpha = 2 \arctan\left(- \frac 1 2 + \sqrt{\sqrt{3}-\frac 3 4}\right) .
Now we can plug this into E=(cosα,sinα) E = (\cos \alpha, \sin \alpha) and get E=(34+12334,12+143+12394) 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 EC as 6+3+12397+23+433 \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 C which we subtract from 18060 180^\circ - 60^\circ to get our final answer of 2π3+arctan(94312814+123)37.9235 \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.

Henry U - 1 year ago

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I got the same answer. My solution is above

Steven Chase - 1 year ago

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