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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 =108o and O is the centre of the circle, find :

(i) angle BCT

(ii) angle DOC

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Solution


BC=CD (given)

BCG=108°
Construction: B and D are joined.OC and OD are joined.

To find: ∠DOC and ∠BCT

as BCG+BCD=180° (linear pair)

So
angleBCD=180°108°=72°

CBD=CDB=x°(as BC=CD, △BCD is isosceles, so two angles are equal)
In △BCD,
CBD+CDB+BCD=180°

x+x+72°=180°
2x=180−72

x=54°
and BCT=CDB (angle −tangent theorem)

as
CDB=54°
so
BCT=54°& CBD=54°

DOC=2×CBD( as angle at the center is 2 times of angle at cicumference on the same chord)

so
DOC=2×54°=108°
Hence DOC=108° and

BCT=54°


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