In the figure O is the centre of the circle and angle AOB= 80° .Find the value of
☆angle AOC
☆ angle OAC
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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
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
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Answered by
3
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 = = 40°
- So, ∠AOC = ∠OAC = 40° (∵ OA = OC)
∴ ∠AOC = ∠OAC = 40°
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