Math, asked by sakshiyadav279, 9 months ago

ABCD is a cyclic quadrilateral in
which AB = AD. angle BCD = 70°.
Find (i) m (arc BCD) (ii) m (arc BAD)
(iii) angle ADB.​

Answers

Answered by amitnrw
29

m (arc BCD) =220° , m (arc BAD) = 140° , ∠ADB= 35°

Step-by-step explanation:

ABCD is a cyclic quadrilateral in

which AB = AD

∠BCD =  70°

angle by BD at center  = 2 * 70° = 140°

=> m (arc BCD) = 360° - 140° = 220°

m (arc BAD) = 360° - 220° = 140°

∠BCD =  70°

∠BAD + ∠BCD = 180°  ( cyclic quadrilateral)

=> ∠BAD + 70° = 180°

=> ∠BAD = 110°

in Δ ADB

AB = AD

=> ∠ADB = ∠ABD

∠BAD + ∠ADB + ∠ABD= 180°

=> 110° + ∠ADB + ∠ADB= 180°

=> 2∠ADB= 70°

=> ∠ADB= 35°

Learn more:

ABCD एक चतुर्भुज है जिसमें AB=AD और BD=CD और DBC=2 ...

https://brainly.in/question/15266081

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In a circle with centre o an arc abc subtends an angle of 110 degree at

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Answered by meghana9715
19

Hope this helps you

Thank you

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