Can someone solve this geometry/algebra problem for me
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Can someone solve this geometry/algebra problem for me

[From: ] [author: ] [Date: 12-05-23] [Hit: ]
Its irrelevant, though true, so you can disregard it.The measure of the arc going from B to A without passing through C (the arc going the other way around the circle, if that makes any sense.Ill call it arc y to simplify stuff) + the measure of arc ACB must be 360°.......
Angle ACB is an inscribed angle of circle P. Find x if angle ACB measures x° and arc ACB measures (x+45)°

I am really stuck on this and have no idea where to begin to solve this. It would be great if someone could help me! :)

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Ok. This is hard to describe purely with text, but I'll try. I recommend trying to picture this, or even drawing it to help you get it.

[I'll call the center p, since it's circle P. Angle ApB must equal the same as the length of the arc since it is composed of 2 radii. Thus, angle ApB equals (x+45)° . ] - I orginally thought I needed this. It's irrelevant, though true, so you can disregard it.

The measure of the arc going from B to A without passing through C (the arc going the other way around the circle, if that makes any sense. I'll call it "arc y" to simplify stuff) + the measure of arc ACB must be 360°. m stands for measure. So, m(arc)(y)+m(arc)(ACB)=360, and thus m(arc)(y)+x+45°=360°. So, m(arc)(y)=(315-x)°.

You must also know that the measure of an angle inscribed in a circle is half of the measure of the arc it intercepts. Thus, m∠ACB is half of the arc(y). If ACB is x°, then arc(y) must be (2x)°.

Taking both of those, m(arc)(y)=(315-x)° and m(arc)(y)=(2x)°.
Thus, 315-x=2x.

Solve for x!
315=3x
x=105.

(That was a doozy... took a loooong time)
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