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\(\overline{BC}\) is a diameter of circle O and is parallel to \(\overline{YZ}\). If the length of arc ABC is 22π and the length of minor arc CD is 11π, then what is the positive difference between the degree measures of \(∠ OYZ\) and \(∠ YOZ\)?
A. 30
B. 45
C. 60
D. 80
E. 90
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yashikaaggarwal
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Since YZ is parallel to BC
ZO is transversal
Angle YZO = Angle BOA (alternate angles) = 40
Arc DA is half of arc ABC
so angle DOA is half of angle AOC
Angle AOC = Angle BOC + Angle AOB
Angle AOC = 180+40
Angle AOC = 220
Angle DOA = 220/2 = 110
Triangle YOZ
Angle YZO + Angle ZYO + Angle YOZ = 180
40+ Angle ZYO + 110 = 180
Angle ZYO = 30
Difference between degree measures of ∠ZYO and ∠YOZ
=> 110 - 30 = 80
Answer is D
_________________
ZO is transversal
Angle YZO = Angle BOA (alternate angles) = 40
Arc DA is half of arc ABC
so angle DOA is half of angle AOC
Angle AOC = Angle BOC + Angle AOB
Angle AOC = 180+40
Angle AOC = 220
Angle DOA = 220/2 = 110
Triangle YOZ
Angle YZO + Angle ZYO + Angle YOZ = 180
40+ Angle ZYO + 110 = 180
Angle ZYO = 30
Difference between degree measures of ∠ZYO and ∠YOZ
=> 110 - 30 = 80
Answer is D
_________________
Posts: 4,849
\(\angle{YZO} = \angle{ZOB} = 40^o\) (Alternate angles between 2 parallel lines are equal)
\(Arc \space \angle{ABC} = \angle{AOB} + \angle{BOC} = 40 + 180 = 220^o\)
Length of arc ABC = \(22\pi = \frac{\theta}{360} * 2 \pi r\)
\(22\pi = \frac{220}{360} * 2 \pi r\)
Solving we get r = 18
Length of arc CD = \(11\pi = \frac{\theta}{360} * 2 \pi r\)
\(11\pi = \frac{\theta}{360} * 2 \pi * 18\)
Solving, we get \(\theta = 110^o\)
\(\angle{YOZ} = 180 - (\angle{AOB} + \angle{COD}) = 180 - (220 + 110) = 30^o\). (Sum of angles on a straight line = 180)
\(\angle{OYZ} = 180 - (\angle{YOZ} + \angle{YZO}) = 180 - (30 + 40) = 110^o\). (Sum of angles of a triangle = 180)
Difference = 110 - 30 = 80
Option D
Arun Kumar
\(Arc \space \angle{ABC} = \angle{AOB} + \angle{BOC} = 40 + 180 = 220^o\)
Length of arc ABC = \(22\pi = \frac{\theta}{360} * 2 \pi r\)
\(22\pi = \frac{220}{360} * 2 \pi r\)
Solving we get r = 18
Length of arc CD = \(11\pi = \frac{\theta}{360} * 2 \pi r\)
\(11\pi = \frac{\theta}{360} * 2 \pi * 18\)
Solving, we get \(\theta = 110^o\)
\(\angle{YOZ} = 180 - (\angle{AOB} + \angle{COD}) = 180 - (220 + 110) = 30^o\). (Sum of angles on a straight line = 180)
\(\angle{OYZ} = 180 - (\angle{YOZ} + \angle{YZO}) = 180 - (30 + 40) = 110^o\). (Sum of angles of a triangle = 180)
Difference = 110 - 30 = 80
Option D
Arun Kumar
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