1. ## Arc Length help!

I have a word problem im not sure how to solve regarding arc length

A radius of 93,000,000 miles measured between the earth and the sun at a central angle measuring 31 feet. Find the diameter of the sun(arc length in this case)

How do I solve with the central angle measured in feet?

2. Originally Posted by rusty1363
I have a word problem im not sure how to solve regarding arc length

A radius of 93,000,000 miles measured between the earth and the sun at a central angle measuring 31 feet. Find the diameter of the sun(arc length in this case)

How do I solve with the central angle measured in feet?
central angle from where? is there a diagram given?

it might help if you post the problem exactly as it is written.

3. yes there is a picture but ill write the text verbatim: At a time when the earth was 93,000,000 miles from the sun, you observe through a tinted glass that the diameter of the sun occupied an arc of 31'. Determine to the nearest ten thousand miles the diameter of the sun. Hint: Because the radius of arc AB is large and the central angle is small, the length of the diamater of the sun is approx. the length of the arc AB.

So the central angle is measured from the earth from the two radii that form the arc AB (the diameter of the sun).

4. Originally Posted by rusty1363
yes there is a picture but ill write the text verbatim: At a time when the earth was 93,000,000 miles from the sun, you observe through a tinted glass that the diameter of the sun occupied an arc of 31'. Determine to the nearest ten thousand miles the diameter of the sun. Hint: Because the radius of arc AB is large and the central angle is small, the length of the diamater of the sun is approx. the length of the arc AB.

So the central angle is measured from the earth from the two radii that form the arc AB (the diameter of the sun).
arc length formula ...

$\displaystyle s = r \cdot \theta$

$\displaystyle r$ = 93 million miles

$\displaystyle \theta$ = central angle in radians

fyi, ... 31' = 31 minutes of arc, not feet

you'll have to convert 31' to radians for $\displaystyle \theta$

5. ok thanks!