Q. 20

In the given figure, O is the centre of a circle. If AOB = 100° and AOC = 90°, then BAC = ?


A. 85o

B. 80o

C. 95o

D. 75o

Answer :

Given: AOB = 100° and AOC = 90°,


Consider ΔOAB


Here,


OA = OB (radius)


Let OBA = OAB = x (angles opposite to equal sides are equal)


By angle sum property


AOB + OBA + OAB = 180°


100° + x + x = 180°


2x = 180° – 100°


2x = 80°


x = 40°


Similarly, in ΔAOC


OA = OC (radius)


Let OCA = OAC = y (angles opposite to equal sides are equal)


By angle sum property


AOC + OCA + OAC = 180°


90° + y + y = 180°


2y = 180° – 90°


2y = 90°


y = 45°


Here,


BAC = OAB + OAC = x + y = 40° + 45° = 85°


BAC = 85°

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