Math, asked by renarasheed5151, 2 months ago

In the given figure AD is a diameter of a circle. If <BCD = 140°, calculate <BAD and < ADB
please find me the answer ☺️​

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Answered by Anonymous
1

Answer:

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Answered by Anonymous
0

Answer:

(i) join BD

now,ABCD is a cyclic quadrilateral

angle BAD +angle BCD =180°( oppo. angle of a cyclic quadrilateral are equal)  

angle BAD+150°=180°  

angle BAD =180°-150°  

=30°

(ii) angle ABD =90°(angle in a semi -circle )

now,in triangle ABD ,we have  

angle ABD +angle BAD +angle ADB =180°  

90°+30°+ angle ADB =180°angle ADB =180°-120°  

= 60°

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