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Question
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
Solution
In cyclic quadrilateral ABCD,
∠B + ∠D = 180° (opp angles of the cyclic quad are supplementary)
⇒ 100° + ∠ADC = 180°
=> ∠ADC = 80°
Now in ΔACD
∠ACD + ∠CAD + ∠ADC = 180°
40° + ∠CAD + 80° = 180°
∠CAD = 180° - 120° = 60°
Now ∠DCT = ∠CAD (angles in the alternate segment)
∴ ∠DCT = 60°
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