Properties of circle(moving circle). part 3

Geometry Level 2

In the above figure, A and B are centres of the circles shown. The circles intersect at C and D .Find ∠ C E D \angle CED + ∠ C F D \angle CFD ( in degree )


The answer is 120.

Gaurav Singh by Gaurav singh , 21, India

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2 solutions

Marta Reece
Jun 12, 2017

△ A B C \triangle ABC is equilateral since all of its sides are r r .

That makes ∠ A B C = 6 0 ∘ \angle ABC=60^\circ

∠ D B C = 2 × ∠ A B C = 12 0 ∘ \angle DBC=2\times\angle ABC=120^\circ

∠ D F C = 1 2 ∠ D B C = 6 0 ∘ \angle DFC=\dfrac12\angle DBC=60^\circ

∠ C E D + ∠ D F C = 2 × 6 0 ∘ = 12 0 ∘ \angle CED+\angle DFC=2\times60^\circ=\boxed{120^\circ}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Equilateral triangle

0 Helpful 0 Interesting 0 Brilliant 0 Confused

The asker must add to his question that we're searching for degrees, not radians, since we can't be sure otherwise.

mathh mathh - 6 years, 11 months ago

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thanks bro for suggestion.

Gaurav singh - 6 years, 10 months ago

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