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Somebody help me

 Mar 1, 2021
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To be honest, I do not recall the correct way to do this problem. Of course, that assumes that there is a "correct" way, but I will just show you my method, and I hope that it makes sense to you! I also created a diagram. Admittedly, this was probably not necessary, but I did it anyway.

 

 

OC = OA = OB because those lengths represent radii of the circle O, and the radius of a circle is constant.

Therefore, \(\triangle OCA\) is isosceles, and so is \(\triangle OBA\). Because they are isosceles triangles, the angles opposite those congruent sides have the same measure.

 

Therefore, \(m\angle OCA = m\angle OCB = 18^\circ\)

By angle addition, \(m\angle ACB = 36^\circ\).

 

Now, figure \(ACBO\) is a quadrilateral, so that means that the sum of the interior angles is 360. We already know 3 of the angles, so let's find the 4th and final angle.

 

\(\text{4th interior angle } = 360 - (18 + 18 + 36)\\ \text{4th interior angle } = 360 - 72\\ \text{4th interior angle } = 288\)

 

\(m\angle AOB = 360 - \text{4th interior angle}\\ m\angle AOB = 360 - 288\\ m\angle AOB = 72^\circ\)

 

I hoped this helped you on your solving adventures!

 Mar 1, 2021

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