Mathematics
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠BCG = 108° and O is the center of the circle, find :
(i) angle BCT
(ii) angle DOC
Circles
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Answer
Join OC, OD and BD.
Given,
∠BCG = 108°
From figure,
⇒ ∠BCG + ∠BCD = 180° [Linear pairs]
⇒ 108° + ∠BCD = 180°
⇒ ∠BCD = 180° - 108°
⇒ ∠BCD = 72°.
From figure,
⇒ ∠BDC = ∠DBC = x(let) [As, angles opposite to equal sides are equal]
In triangle BDC,
⇒ ∠DBC + ∠BDC + ∠BCD = 180° [Angle sum property of triangle]
⇒ x + x + 72° = 180°
⇒ 2x + 72° = 180°
⇒ 2x = 180° - 72°
⇒ 2x = 108°
⇒ x =
⇒ x = 54°.
From figure.
∠BCT = ∠BDC (Angles in alternate segment are equal)
∠BCT = 54°.
Hence, ∠BCT = 54°.
(ii) As, angle subtended by a segment on center is twice the angle subtended by it on any other part of circumference.
⇒ ∠DOC = 2∠DBC
⇒ ∠DOC = 2(54°) = 108°.
Hence, ∠DOC = 108°.
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