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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