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Mathematics 10 Online
OpenStudy (anonymous):

The central angle of an arc measures 12°. The radius of its circle is 30 inches. What is the length of the arc to the nearest quarter inch

OpenStudy (anonymous):

1 in 6 1/4 in 37 1/2 in 94 1/4 in

OpenStudy (anonymous):

@Owlcoffee

OpenStudy (owlcoffee):

A central angle has the quality that is is equal the arc it intersects. The definition of a central angle is that it's an angle, composed of two radiuses and the center of the circumference and is equal the arc it intersects. |dw:1417916400648:dw| \[<O=arc.AB\] now, what we have to do, knowing that, is calculate the perimeter of the circle and sustract 12 from it.

OpenStudy (anonymous):

The perimeter of the circle ?

OpenStudy (anonymous):

@Owlcoffee

OpenStudy (owlcoffee):

yes, it's calculated: \[P_\circ =2 \pi r\]

OpenStudy (owlcoffee):

Does it specify the what arc they want you to calculate?

jimthompson5910 (jim_thompson5910):

You find the perimeter but you don't subtract 12 from it. You find the perimeter and then multiply that perimeter (aka circumference) by 12/360

OpenStudy (owlcoffee):

Yes, I was doubting that method would work out. thanks @jim_thompson5910

OpenStudy (anonymous):

30

OpenStudy (anonymous):

@Owlcoffee

OpenStudy (anonymous):

or 94.2?

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

@GodlyMaths

jimthompson5910 (jim_thompson5910):

what circumference did you get?

OpenStudy (anonymous):

6.28

jimthompson5910 (jim_thompson5910):

use C = 2*pi*r where r = 30 in this case

OpenStudy (anonymous):

188.4

jimthompson5910 (jim_thompson5910):

now multiply that by 12/360

jimthompson5910 (jim_thompson5910):

you are doing this because the arc is 12/360 of the circle's full perimeter

OpenStudy (anonymous):

okay so i multiply 188.4 x 12 x 360 ? or 188.4 x 12/360?

jimthompson5910 (jim_thompson5910):

188.4 x 12/360

OpenStudy (anonymous):

6.28

jimthompson5910 (jim_thompson5910):

round that to the nearest quarter inch

OpenStudy (anonymous):

6 1/4?

jimthompson5910 (jim_thompson5910):

good

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