20. O is the centre of the circle such that OM Perpendicular AB. If angle ABC is 42 degree, find angle AOC and angle ODC. Hence prove that ADCO is a cyclic quadrilateral.
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Asked on October 15, 2019 by
Chavi Saluja
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.
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ANSWER
Given:∠ABC=100
o
We know that, ∠ABC+∠ADC=180
o
(The sum of opposite angles in a cyclic quadrilateral =180
o
)
∴100
o
+∠ADC=180
o
∠ADC=180
o
−100
o
∠ADC=80
o
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
o
=160
o
In ΔAOC,
∠AOC+∠OAC+∠OCA=180
o
160
o
+∠OCA+∠OCA=180
o
[∵∠OAC=∠OCA]
2∠OCA=20
o
∠OCA=10
o
∠OCA+∠OCD=40
o
10
o
+∠OCD=40
o
∴ ∠OCD=30
o
Hence, ∠OCD+∠DCT=∠OCT
∵∠OCT=90
o
(The tangent at a point to circle is ⊥ to the radius through the point to constant)
30
o
+∠DCT=90
o
∴ ∠DCT=60
o
.