In the given figure AD is a diameter of a circle. If <BCD = 140°, calculate <BAD and < ADB
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(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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