Question

In the given circle with the centre O,

angle ABC = 100° angleACD = 40° and CT is a tangent to the circle at C. Find Angle ADC and Angle DCT.
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Answer 1

See below ~

Explanation:

∠ADC

• ∠ABC + ∠ADC = 180° (Angles in a cyclic quadrilateral)
• ∠ADC = 180° - 100° = 80°

∠DCT

• ∠DCT = ∠CAD (ANGLES IN ALTERNATE SEGMENTS EQUAL)
• ∠CAD = 180° - 80° - 40° = 60°
• ∠DCT = 60°

Answer 2

∠ADC = 80° and ∠DCT = 60°

Explanation:

As the points A, B, C and D lie on a circle, So, ABCD is a cyclic quadrilateral...

As, Sum of opp. ∠s in a cyclic quadrilateral = 180°

∴ ∠ADC + ∠ABC = 180°

=> ∠ADC + 100° = 180° => ∠ADC = 80°

Now,

Sum, of all angles in a traingle = 180°

In ΔACD, ∠CAD + ∠ADC + ∠ACD = 180°

=> ∠CAD + 80° + 40° = 180°

=> ∠CAD = 180° - 80° - 40° = 60°

As CT is tangent to the circle at the point C and CD is a chord of the circle, So

∠DCT = ∠CAD........(∠s in alternate segments are equal,,So ∠CAD = 60°)

=> ∠DCT = 60°

Hope it helps you!!