Question

in the given circle, O is the centre. Chords AD = CD, if Z ABC = 40 , find Z ABD , Z BOC and Z COD

Answer 1

Step-by-step explanation:

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

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∠OCA=10

o

∠OCA+∠OCD=40

o

10

o

+∠OCD=40

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∴ ∠OCD=30

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Hence, ∠OCD+∠DCT=∠OCT

∵∠OCT=90

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(The tangent at a point to circle is ⊥ to the radius through the point to constant)

30

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+∠DCT=90

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∴ ∠DCT=60

o

.