Q. 20

# In the given figu

Given: AOB = 100° and AOC = 90°,

Consider ΔOAB

Here,

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

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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