Question

EXERCISE 12B
1. (i) In Figure (1), O is the centre of the circle. If
ZOAB = 40° and ZOCB = 30°, find ZAOC.
(ii) In Figure (2), A, B and Care
А
three points on the circle with
centre O such that ZAOB = 90°
90°
and ZAOC = 110°. Find ZBAC.
110°
B
C​

Answer 1

Answer:

Consider △ BOC We know that the sides are equal to the radius So we get OC = OB From the figure we know that the base angles of an isosceles triangle are equal ∠OBC = ∠OCB It is given that ∠OCB = 30o So we get ∠OBC = ∠OCB = 30o So we get ∠OBC = 30o ……. (1) Consider △ BOA We know that the sides are equal to the radius So we get OB = OC From the figure we know that the base angles of an isosceles triangle are equal ∠OAB = ∠OBA It is given that ∠OAB = 40o So we get ∠OAB = ∠OBA = 40o So we get ∠OBA = 40o ……. (2) We know that ∠ABC = ∠OBC + ∠OBA By substituting the values ∠ABC = 30o + 40o So we get ∠ABC = 70o We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference. It can be written as ∠AOC = 2 × ∠ABC So we get ∠AOC = 2 × 70o By multiplication ∠AOC = 140o Therefore, ∠AOC = 140oRead more on Sarthaks.com - https://www.sarthaks.com/727705/in-figure-o-is-the-centre-of-the-circle-if-oab-40and-ocb-30-find-aoc