Mathematics
In the adjoining figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to
50°
40°
60°
70°
Circles
8 Likes
Answer
From figure,
OA = OB (Radius of the circle.)
So, △OAB is an isosceles triangle with ∠OBA = ∠OAB (As angles opposite to equal sides are equal.)
∠OBA = 40°.
Since, sum of angles in a triangle = 180°.
⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 40° + 40° + ∠AOB = 180°
⇒ 80° + ∠AOB = 180°
⇒ ∠AOB = 180° - 80°
⇒ ∠AOB = 100°.
Arc AB subtends ∠AOB at centre and ∠ACB at remaining part of circle.
∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of circle.)
100° = 2∠ACB
∠ACB = = 50°.
Hence, Option 1 is the correct option.
Answered By
5 Likes
Related Questions
In the figure (i) given below, AB is a chord of the circle with centre O, BT is tangent to the circle. If ∠OAB = 32°, find the values of x and y.
In the figure (ii) given below, O and O' are centres of two circles touching each other externally at the point P. The common tangent at P meets a direct common tangent AB at M. Prove that,
(i) M bisects AB.
(ii) ∠APB = 90°.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to
80°
50°
40°
30°
In the adjoining figure, O is the centre of the circle. If ∠BAO = 60°, then ∠ADC is equal to
30°
45°
60°
120°
