Question

A quadrilateral ABCD is inscribed in a circle with centre O. If ∠BOC = 92° and ∠ADC = 112°, then ∠ABO is equal to: Options: ​

Answer 1

Step-by-step explanation:

Given A quadrilateral ABCD is inscribed in a circle with centre O. If ∠BOC = 92° and ∠ADC = 112°, then ∠ABO is equal to:

• We know that sum of opposite angles of a cyclic quadrilateral is equal to 180 degree.
• Angle ADC + angle ABC = 180
• Substituting the given values we get
• 112 + angle ABC = 180
• Angle ABC = 180 – 112
•                 = 68 degree
• Now in triangle BOC, OC = OB
• Here radius of circle are equal.
• Let angle BCO = angle CBO = a
• Angle BOC = 92 degree (given)
• Therefore, angle BOC + angle CBO = angle ABC
• Substituting the values, we get
• 92 + a + a = 180
• 2 a = 180 – 92  
• 2 a = 88
• Or a = 44 degree
• Now angle ABO + angle CBO = angle ABC  quality
• Substituting we get
• Angle ABO + 44 = 68
• Angle ABO = 68 – 44
• Angle ABO = 24 degree.

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