# 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 $O$ is the center of the circle. If you only knows measure of $\overline{OA},\overline{OB}, \overline{OC}$ and $\angle ACO$ then can you find measure of $\angle AOB$ ?

Any answer will be worthy and will be appreciate, so please try once! Note by Zakir Husain
1 year, 1 month ago

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

However, I think $OA$ and $OB$ are the same (radii), so only one needs to be specified.

- 1 year, 1 month ago

Yeah, That's right

- 1 year, 1 month ago

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.

- 1 year, 1 month ago

You can do it using Sine rule

$\frac{OB}{sin C}$ = $\frac{OC}{sin OBC}$

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

- 1 year, 1 month ago

- 1 year, 1 month ago

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

- 1 year, 1 month ago

- 1 year, 1 month ago

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

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

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

- 1 year, 1 month ago

@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!

- 1 year, 1 month ago