Problem:
Points A,B and C on a circle of radius r are situated so that AB=AC,AB>r, and the length of minor arc BC is r. If angles are measured in radians, then AB/BC=
Answer Choices:
A. 21csc41
B. 2cos21
C. 4sin21
D. csc21
E. 2sec21
Solution:
Draw and label the figure as shown, where O is the center of the circle. In radians, ∠BOC= r/r=1, so ∠BAC=1/2 and ∠BAE=1/4. Since BE⊥EA, it follows that
BCAB=21BEAB=21csc∠BAE=21csc41.
OR
Since ∠BAC=21, it follows that ∠ACB=21(π−21)=2π−41. The length of a chord subtended by the inscribed angle β in a circle of radius r is 2rsinβ, so
AB=2rsin∠ACB=2rsin(2π−41)=2rcos41BC=2rsin∠BAC=2rsin21=4rsin41cos41
Therefore, BCAB=4rsin41cos412rcos41=2sin411=21csc41.
OR
By the Law of Sines, BCAB=sinAsinC. But 2∠C+∠A=π, so ∠C=2π−2∠A and sinC=cos2A. Since ∠A=21,
BCAB=sinAsinC=2sin2Acos2Acos2A=2sin2A1=21csc2A=21csc41.
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