SOLUTION: the central angle theta of a circle with radius 9 inches intercepts an arc 20 inches, find theta

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Question 378973: the central angle theta of a circle with radius 9 inches intercepts an arc 20 inches, find theta
Found 2 solutions by jsmallt9, robertb:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This problem can be solved with a simple proportion:
          Angles              Arcs         

Part:     central angle          intercepted arc
          -----------------  =  -------------------
Total:    full circle (360)      full circumference

(Since Algebra.com's software doesn't "do" theta, I am going to use q in place of theta. So replace any q's below with theta.)
Your proportion is, therefore:
q%2F360+=+20%2F%282%2A9%2Api%29 (Since Circumference = 2*r*pi.)
This simplifies as follows:
q%2F360+=+10%2F%289%2Api%29
To solve for q we start by cross-multiplying:
q%2A9pi+=+360%2A10
which simplifies to:
q%2A9pi+=+3600
Next we divide by 9pi
q+=+3600%2F%289pi%29
The fraction reduces:
q+=+400%2Fpi
So q (or theta) is 400%2Fpi degrees.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
From the (circular) arc length formula, s+=+r%2Atheta, use it to solve for theta:
theta+=+s%2Fr+=+20%2F9+radians. In degrees, %2820%2F9%29%2A%28180%2Fpi%29+=+400%2Fpi degrees.