Question

In the given figure, O is the centre of a circle. If ∠OAB = 40° and C is a point on the circle, then ∠ACB =

(a) 40°
(b) 50°
(c) 80°
(d) 100°​

Answer 1

Answer: (D) 100°

OA=OB (RADII)

∠OAB=∠OBA ( Angles opposite to equal sides of a triangle are equal)

∠OBA= 40°

IN ΔOAB

∠AOB+∠OAB+∠OBA=180° (ANGLE SUM PROPERTY)

40°+40°+∠AOB=180°

∠AOB=180°-80°

∠AOB=100°

STAN KPOP IDOLS

✌

Answer 2

Answer:

ANSWER

∵OA=OB (radius of circle)

∴∠OAB=∠OBA (angles opposite to equal sites)

∠OBA=40

0

In right angled ΔOAB

∠OAB+∠OBA+∠AOB=180

0

40

0

+40

0

+∠AOB=180

0

∠AOB=180

0

–80

0

∠AOB=100

0

We know that

∠ACB=

2

1

∠AOB

=

2

1

×100

0

=50

0

Thus, (A) is correct