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Question
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 centre of the circle, find :
- angle BCT
- angle DOC
Solution
Join OC, OD and AC
i. ∠BCG + ∠BCD = 180° ...(Linear pair)
`=>` 108° + ∠BCD = 180° ...(∵ ∠BCG = 108° given)
`=>` ∠BCD = 180° – 108° = 72°
BC = CD ...(Given)
∴ ∠DCP = ∠BCT
But, ∠BCT + ∠BCD + ∠DCP = 180°
∴ ∠BCT + ∠BCT + 72° = 180° ...(∵ ∠DCP = ∠BCT)
2∠BCT = 180° – 72° = 108°
∴ ∠BCT = `108^circ/2` = 54°
ii. PCT is a tangent and CA is a chord.
∠CAD = ∠BCT = 54°
But arc DC subtends ∠DOC at the centre and ∠CAD at the remaining part of the circle.
∴ ∠DOC = 2∠CAD
= 2 × 54°
= 108°
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