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°
Rate this question :






















Prove that there is one and only one circle passing through three non – collinear points.
RS Aggarwal & V Aggarwal - MathematicsIn the adjoining figure, chords and
of a circle with center
intersect at right angles at
If
calculate
Number of circles that can be drawn through three non-collinear points is
RD Sharma - MathematicsIn the given figure, is an isosceles triangle in which
and a circle passing through
and
intersects
and
at
and
respectively.
Prove that
In the adjoining figure, and
are two equal chords of a circle with center
Show that
lies on the bisector of
In the adjoining figure, two circles with centers at and
and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of meets the bigger circle in
and
find the length of
In the given figure, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 50°. Then, ∠AOC = ?
The question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer:
In the given figure, O is the centre of a circle and ∠AOC = 120°. Then, ∠BDC = ?