Question
Download Solution PDFA circle is circum-scribed of ΔABC. O is the centre of the circle and CP is the tangent of the circle at the point C. If BC bisects the ∠OCP, then what is the supplementary angle of ∠BAC?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFAs we know,
⇒ ∠OCP = 90° (Tangent)
⇒ BC bisects ∠OCP
∠OCB = ∠OBC = 45° [OB = OC]
In ΔBOC,
⇒ ∠OCB + ∠OBC + ∠BOC = 180°
⇒ ∠BOC + 45° + 45° = 180°
⇒ ∠BOC = 180° – 90° = 90°
⇒ ∠BAC = 1/2 × ∠BOC = (1/2) × 90° = 45°
As we know,
Sum of two supplementary angle is 180°.
Let the supplementary angle of ∠BAC be x, then
⇒ x + ∠BAC = 180°
⇒ x = 180° – 45° = 135°
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