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I have found a question like following:

enter image description here

Its asked that what could be the angle $x$ if $BC$ is not diameter of the circle.

So, my question is if it possible to be greater then $90^{\circ}$ for an angle like $x$?

My proof said:

The highest inscribed angle only could be made of with the chord same diameter. The other chords could not create an angle greater than that.

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    $\begingroup$ The picture certainly makes it look possible. The angle can be made arbitrarily close to $180^\circ$ by making $BC$ short enough. $\endgroup$ Aug 30 '14 at 6:16
  • $\begingroup$ the picture is figure to scale, I need the theory. The chord could be anywhere. $\endgroup$ Aug 30 '14 at 6:28
  • $\begingroup$ For given radius $r$, the angle depends only on the length of $BC$, and is independent of the location of $A$, in the sense that if you take an $A'$ on the same side of $BC$ as $A$ is the angle will still be $x$. Make the triangle ABC isosceles, and you can use trig to compute explicitly. $\endgroup$ Aug 30 '14 at 6:32
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Yes, the angle can get very close to (but not exactly equal to) 180 degrees. For every angle value greater than 90-degrees the chord $\overline{BC}$ is smaller than the diameter.

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  • $\begingroup$ For every angle value less than 90-degrees the chord is smaller than the diameter. Is that possible to have a chord greater then diameter? $\endgroup$ Aug 30 '14 at 6:33
  • $\begingroup$ @KhandakerMustakimurRahman I meant to say, for every angle greater than 90-degrees the chord $\overline{BC}$ is smaller than the diameter. No, it's not possible to have a chord larger than the diameter. $\endgroup$
    – Addem
    Aug 30 '14 at 6:36
  • $\begingroup$ I am sorry, I could't convince with your answer. Could you provide any solid proof of it, or any type of link from where I could get it. $\endgroup$ Aug 30 '14 at 6:44
  • $\begingroup$ @KhandakerMustakimurRahman Proof: Take the unit circle which has formula $x^2+y^2=1$. Clearly the point $(1,0)$ is on the circle and so is $(3/5, 4/5)$ and so is $(3/5, -4/5)$. Find the vectors coming from $(1,0)$ to these other points. They are $\vec{v}=\langle -2/5, 4/5\rangle, \vec{w}=\langle -2/5, -4/5\rangle$ and then use the angle formula for a dot-product. $\vec{v}\cdot\vec{w}=|\vec{v}|\cdot |\vec{w}|\cos\theta$. In this case that means $4/25 + 16/25 = 2\cdot \sqrt{4/25+16/25}\cos \theta$. Solving for theta you should find it's greater than 90 degrees. $\endgroup$
    – Addem
    Aug 30 '14 at 6:51
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Note that you can also put a point $A'$ on the other side of the chord (i.e. on the longer arc of the circle), which would give an acute angle.

Using the theorem that the angle at the centre is twice the angle at the circumference (which you may know) you will find the angles $A+A'= 180^{\circ}$ - because the total angle at the centre is $360^{\circ}$. So if the chord is not a diameter, one of the angles will be obtuse and the other acute.

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