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Difficulty: HardCircle Geometry

In a circle with center OO, ABAB is a diameter. Point CC lies on the circle such that the measure of arc ACAC is 5π9\frac{5\pi}{9} radians. Point DD lies on the circle such that chord ACAC is parallel to segment ODOD, and points CC and DD lie on the same side of diameter ABAB. What is the measure, in radians, of angle CODCOD?

  1. A
    5π18\frac{5\pi}{18}
  2. 2π9\frac{2\pi}{9}Answer
  3. C
    4π9\frac{4\pi}{9}
  4. D
    13π18\frac{13\pi}{18}

Answer

The correct answer is 2π9\frac{2\pi}{9} radians.
The correct answer is 2π9\frac{2\pi}{9} radians. First, the central angle ∠AOC\angle AOC has a measure of 5π9\frac{5\pi}{9} radians because it subtends an arc of the same measure. Since OAOA and OCOC are both radii of the circle, triangle AOCAOC is isosceles with OA=OCOA = OC, which means the base angles are equal: ∠OAC=∠OCA=π−5π/92=2π9\angle OAC = \angle OCA = \frac{\pi - 5\pi/9}{2} = \frac{2\pi}{9} radians. Since chord ACAC is parallel to segment ODOD and the diameter ABAB acts as a transversal line, the corresponding angles ∠OAC\angle OAC and ∠BOD\angle BOD are equal, so ∠BOD=2π9\angle BOD = \frac{2\pi}{9} radians. Finally, because AA, OO, and BB form a straight line, the angles along the diameter must sum to π\pi radians: ∠COD=π−∠AOC−∠BOD=π−5π9−2π9=2π9\angle COD = \pi - \angle AOC - \angle BOD = \pi - \frac{5\pi}{9} - \frac{2\pi}{9} = \frac{2\pi}{9} radians.

Step-by-Step Solution

1
Identify the measure of the central angle ∠AOC\angle AOC from the given arc measure.
The central angle ∠AOC=5π9\angle AOC = \frac{5\pi}{9} radians.
The measure of an arc in radians is equal to the measure of its subtended central angle.
2
Determine the measure of the inscribed angle ∠OAC\angle OAC using the properties of triangle AOCAOC.
∠OAC=2π9\angle OAC = \frac{2\pi}{9} radians.
Since OAOA and OCOC are radii, triangle AOCAOC is isosceles with OA=OCOA = OC, meaning ∠OAC=∠OCA\angle OAC = \angle OCA. The sum of angles in a triangle is π\pi radians, so ∠OAC=π−∠AOC2=π−5π92=2π9\angle OAC = \frac{\pi - \angle AOC}{2} = \frac{\pi - \frac{5\pi}{9}}{2} = \frac{2\pi}{9} radians.
3
Use the parallel lines ACAC and ODOD to find the measure of angle ∠BOD\angle BOD.
∠BOD=2π9\angle BOD = \frac{2\pi}{9} radians.
Since chord ACAC is parallel to segment ODOD and diameter ABAB is a transversal line, the corresponding angles ∠OAC\angle OAC and ∠BOD\angle BOD are equal.
4
Calculate the measure of angle CODCOD using the angles along the diameter ABAB.
∠COD=2π9\angle COD = \frac{2\pi}{9} radians.
Points AA, OO, and BB lie on a straight line, so the angles ∠AOC\angle AOC, ∠COD\angle COD, and ∠BOD\angle BOD must sum to π\pi radians. Therefore, ∠COD=π−∠AOC−∠BOD=π−5π9−2π9=2π9\angle COD = \pi - \angle AOC - \angle BOD = \pi - \frac{5\pi}{9} - \frac{2\pi}{9} = \frac{2\pi}{9} radians.

Key Concept

Angle relationships in circles, including central angles, inscribed angles in isosceles triangles, and parallel line transversal properties.
Estimated Time:2m 30s
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Topics

Circle Geometry

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