Given circle A, with BC = 5, DC = 12, and right angle BCD.?

Question

http://i40.photobucket.com/albums/e215/xxs3xp1stolxx/3.jpg

Answer 1

Since BD is a diameter, triangle BCD is a right triangle. By the Pythagorean Theorem

(BD)² = (BC)² + (CD)² = 5² + 12² = 25 + 144 = 169
BD = √169 = 13

2r = BD = 13
r = 13/2
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Let
θ = m
sin(θ/2) = [(1/2)BC] / AB = (1/2)(5) / (13/2) = 5/13
θ/2 = arcsin(5/13)
θ = 2arcsin(5/13) ≈ 0.7895822 radians

The measure in radians of arc BC is θ.
m arcBC = θ ≈ 0.7895822 radians

length arcBC = rθ = (13/2) * 0.7895822 ≈ 5.1322846

m
m