Question
In the adjoining figure 'O' is the center of circle, ∠CAO = 25∘ and ∠CBO = 35∘. What is the value of ∠AOB?
Answer: Option C
:
C
In ΔAOC,
OA=OC --------(radii of the same circle)
∴ΔAOC is an isosceles triangle
→∠OAC=∠OCA=25∘----- (base angles of an isosceles triangle )
In ΔBOC,
OB=OC --------(radii of the same circle)
∴ΔBOC is an isosceles triangle
→∠OBC=∠OCB=35∘ -----(base angles of an isosceles triangle )
∠ACB=25∘+35∘=60∘
∠AOB=2×∠ACB ----(angle at the center is twice the angle at the circumference)
= 2×60∘
=120∘
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:
C
In ΔAOC,
OA=OC --------(radii of the same circle)
∴ΔAOC is an isosceles triangle
→∠OAC=∠OCA=25∘----- (base angles of an isosceles triangle )
In ΔBOC,
OB=OC --------(radii of the same circle)
∴ΔBOC is an isosceles triangle
→∠OBC=∠OCB=35∘ -----(base angles of an isosceles triangle )
∠ACB=25∘+35∘=60∘
∠AOB=2×∠ACB ----(angle at the center is twice the angle at the circumference)
= 2×60∘
=120∘
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