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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
Given
`{:(∠ABC = 100^circ),(∠ACD = 40^circ):}}` ...(Given)
∠ABC + ∠ACD = 180° ...(Opposite angles of a cyclic quadrilateral)
100° + ∠ADC = 180°
∴ ∠ADC = 180° – 100° = 80°
Also,
∠ACD + ∠ADC + ∠CAD = 180° ...(Sum of angles of a triangle)
40° + 80° + ∠CAD = 180°
∠CAD = 180° – 120° = 60°
Now, ∠DCT = ∠CAD = 60° ...(Alternate segment theorem)
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