Question

In the figure O is the centre of the circle and angle AOB= 80° .Find the value of

☆angle AOC
☆ angle OAC

Answer 1

To find this

Angle Aoc +ANGLE AOB =180 (LINEAR PAIR)

ANGLE AOC +80=180

ANGLE AOC =180-80

ANGLE AOC =100

And I don't know about oac sorry
Hope this will useful

Answer 2

Given:

• ∠AOB = 80°

To Find:

• The value ∠AOC
• The value ∠OAC

Solution:

• From the given figure we get to know that OA = OC( radii of the same circle)
• We already know that the base angles of an isosceles triangle are equal.
• In the figure, ∠AOB = 2∠ACB (∵ Angle at the centers is double the angle at the circumference by the same chord).
• So, ∠ACB = $\frac{80}{2}$  = 40°
• So, ∠AOC = ∠OAC = 40° (∵ OA = OC)

∴  ∠AOC = ∠OAC = 40°