C2 Radian measures help Watch

Jamezzy
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So our class has just started a new chapter.

And Ive been stuck on this question for like an hour now its really annoying me.

A minor arc AB of a circle, Centre O and radius 10cm subtends an angle x at O. The major arc AB subtends an angle 5x at O. Find in terms of (Pie), the length of the minor arc AB.

I know the answer but I don't know the method to get to it.

Thanks in advance.
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Danielisew
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(Original post by Jamezzy)
So our class has just started a new chapter.

And Ive been stuck on this question for like an hour now its really annoying me.

A minor arc AB of a circle, Centre O and radius 10cm subtends an angle x at O. The major arc AB subtends an angle 5x at O. Find in terms of (Pie), the length of the minor arc AB.

I know the answer but I don't know the method to get to it.

Thanks in advance.
total circle angle is 6x.
Therefore, considering 360 degrees is two pi, the answer would be 1/6(2 pi) x 10

(as the arc length is the degree multiplied by the radius)
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Jamezzy
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(Original post by Danielisew)
total circle angle is 6x.
Therefore, considering 360 degrees is two pi, the answer would be 1/6(2 pi) x 10

(as the arc length is the degree multiplied by the radius)
Thanks but the answer in the book says 10pi/3 cm.

I was wondering how to get to that from 1/6(2pi) x 10.
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Danielisew
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(Original post by Jamezzy)
Thanks but the answer in the book says 10pi/3 cm.

I was wondering how to get to that from 1/6(2pi) x 10.
1/6 2 pi x 10

(multiply by 2, which makes the 1/6 into 1/3)

= 1/3 pi x 10

(simplify)

= 10pi/3

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rupertj
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(Original post by Jamezzy)
So our class has just started a new chapter.

And Ive been stuck on this question for like an hour now its really annoying me.

A minor arc AB of a circle, Centre O and radius 10cm subtends an angle x at O. The major arc AB subtends an angle 5x at O. Find in terms of (Pie), the length of the minor arc AB.

I know the answer but I don't know the method to get to it.

Thanks in advance.
It is obvious that angle 5x + x = 6x \equiv \ 360^o \equiv \ 2\pi radians.

If you know what 6x is in pi radians, you can find x. Then you use the formula, s = r\theta, where s is the length of the minor arc.
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jimber
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I just literally done this question and came on here for some help for 6B. q. 10 lol
anyway heres what i got:
x = 1/6 x 2pi = 2/6 pi
L = 10 x 2/6 pi
L = 20/6 pi
L = 10/3 pi
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Jamezzy
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(Original post by jimber)
I just literally done this question and came on here for some help for 6B. q. 10 lol
anyway heres what i got:
x = 1/6 x 2pi = 2/6 pi
L = 10 x 2/6 pi
L = 20/6 pi
L = 10/3 pi
lol thanks
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