Question

42. In the given circle with centre 0, angle
ABC = 100°, ACD = 40° and CT is a tangent
to the circle at C. Find ADC and DCT.​

Answer 1

Answer:Given:∠ABC=100 ° We know that, ∠ABC+∠ADC=180 ° (The sum of opposite angles in a cyclic quadrilateral =180=)∴100 ° +∠ADC=180 ° ∠ADC=180 ° −100 ° ∠ADC=80 ° Join OA and OC, we have a isosceles Δ OAC,∵OA=OC (Radii of a circle)∴ ∠AOC=2×∠ADC (by theorem)or ∠AOC=2×80 = =160 ° In ΔAOC, ∠AOC+∠OAC+∠OCA=180 ° 160 ° +∠OCA+∠OCA=180 ° [∵∠OAC=∠OCA]2∠OCA=20 ° ∠OCA=10 = ∠OCA+∠OCD=40 °10 ° +∠OCD=40 ° ∴ ∠OCD=30 ° Hence, ∠OCD+∠DCT=∠OCT∵∠OCT=90°

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