guerrero25 wrote:
A. 8 Pi r /9
B. 18 Pi r /11
C. 11 Pi r /18
D. 9 Pi r /8
E. 4 Pi r /9
The answer is[C] .
From the figure we can calculate the central angle for the arc BC as 50 degree using the fact that (theta/360) * 2*pi*radius = 5 * pi * r / 18.
Then we know that angle CAB = 50/2 = 25 degree. Hence in traingle ABC, we have angle ACB = 180 - (BAC + ABC) = 55 degree. Now the central angle AOB will be twice of the same. Hence AOB = 110 degrees.
Finally using
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