Got stuck at a point -1

While I was spending time on geometry and triangle congruence triangles a question came in my mind which I wasn't able to solve, it can be described as follows:

In the given figure OO is the center of the circle.

If you only knows measure of OA,OB,OC\overline{OA},\overline{OB}, \overline{OC} and ACO\angle ACO then can you find measure of AOB\angle AOB ?

Any answer will be worthy and will be appreciate, so please try once!

Note by Zakir Husain
3 weeks, 5 days ago

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I will try, although I don't know much of geometry(circles and all that).

However, I think OAOA and OBOB are the same (radii), so only one needs to be specified.

Vinayak Srivastava - 3 weeks, 5 days ago

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Yeah, That's right

Zakir Husain - 3 weeks, 5 days ago

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I tried it, but could not do it with the geometry I am taught till now. I think it really requires higher class geometry like the solutions of Aryan Sanghi and Jeff Giif.

Vinayak Srivastava - 3 weeks, 5 days ago

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You can do it using Sine rule

OBsinC\frac{OB}{sin C} = OCsinOBC\frac{OC}{sin OBC}

As we know, OB, Angle C and OC, we can find angle OBC

With this, we can find angle ABO with linear pair

Now, angle A is equal to Angle ABO as OA = OB

So, we can apply sum of anglea of triangles is 180° in triangle AOB to get angle AOB

Aryan Sanghi - 3 weeks, 5 days ago

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Thanks for your solution

Zakir Husain - 3 weeks, 5 days ago

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Yes, I suppose you can! Given sinAa\frac{sinA}{a}=sinBb\frac{sinB}{b}, you can substitute values of angle ACO and side OA, so
sin(angleACO)sideOA=sin(angleOAC)sideOC.\frac{sin(angleACO)}{sideOA}=\frac{sin(angleOAC)}{sideOC}.
Since OA & OB are the radii of a same circle, triangleOAB is an isosceles triangle, so angle AOB =
1802×OAC.180^\circ - 2\times \angle OAC.

Jeff Giff - 3 weeks, 5 days ago

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Thanks for your efforts

Zakir Husain - 3 weeks, 4 days ago

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OAB=OBA\angle OAB^{\circ} = \angle OBA^{\circ}

In AOC\triangle AOC: AOsin(ACO)=OCsin(OAC)    OAC=sin1(OCAO×sin(ACO))\dfrac{AO}{\sin{(ACO^{\circ})}} = \dfrac{OC}{\sin{(\angle OAC^{\circ})}} \implies OAC^{\circ} = \sin^{-1} {\bigg(\dfrac{OC}{AO \times {\sin{(ACO^{\circ})}}}\bigg) \cdot }

BOA=1802(OAC)\begin{aligned} \angle BOA^{\circ} &= 180^{\circ} - 2 (\angle OAC^{\circ}) \end{aligned}

Mahdi Raza - 3 weeks, 4 days ago

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@Zakir Husain, this answer is just a modification from @Jeff Giff's problem. I don't know which of these inspired the other, but both are nice!

Mahdi Raza - 3 weeks, 4 days ago

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