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Jinglander
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Also I am a bit confused about l=C-A/360 * circum can you explain this I dont see it visually
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Thanks sooo much. I am a bit confused what is 2X inscribed in the third step and how does the 360 become 60 in the denominator

Also I am a bit confused about l=C-A/360 * circum can you explain this I dont see it visually

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

So Central Angle =2*Inscribed Angle --> than I substituted Central Angle with 2*Inscribed Angle.

We would have \(L= \frac{2*Inscribed-Angle*2\pi{r}}{360}\) --> reduce by 2*2=4 --> \(L= \frac{Inscribed-Angle*\pi{r}}{90}\) --> \(Inscribed-Angle=\frac{90L}{\pi{r}}\).

About \(L= \frac{Central-Angle}{360}* Circumference\): it describes the fact that the ratio of length of arc to the circumference equals to the ratio of central angle to 360 degrees (the whole circle) --> \(\frac{L}{Circumference}= \frac{Central-Angle}{360}\).

I do advice to check Circles chapter of Math Book (link in my signature) and to revise basics of geometry.

Hope it helps.
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Yes i read the math book on my the app and i didnt see that much detail there. So your varible "inscribed" and "angle" are really the same. I am still confued by Central-Angle/360 shouldn't this just be Central/360 because you said "length of arc to the circumference equals to the ratio of central angle to 360 degrees" why then do you need the subtraction
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Of wait thats a dash not a minus
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Yes i read the math book on my the app and i didnt see that much detail there. So your varible "inscribed" and "angle" are really the same. I am still confued by Central-Angle/360 shouldn't this just be Central/360 because you said "length of arc to the circumference equals to the ratio of central angle to 360 degrees" why then do you need the subtraction

OK:
Inscribed angle when formatted looks like \(Inscribed-Angle\) it's not Inscribed minus angle.
Central angle when formatted looks like \(Central-Angle\) it's not Central minus angle.
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Thanks for all your help. BTW is there a section on here where I can find a good tutor. I am really struggling



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