Q. 20

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

A. 85^{o}

B. 80^{o}

C. 95^{o}

D. 75^{o}

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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