∠AED is formed inside a circle by two intersecting chords. If minor arc BD = 70 and minor arc CA = 180, what is the measure of ∠
AED?
A) 45°
B) 50°
C) 55°
D) 60°
A) 45°
B) 50°
C) 55°
D) 60°
1 answer:
The complete question in the attached figure
we know that
The measure of the interior angle is the half-summit of the arcs that comprise it and its opposite.
so
∠ BED=(1/2)*[arc BD+arc CA]-------> (1/2)*[70+180]----> 125°
∠ BED=125°
∠ CEA=∠ BED-----> 125°
∠ AED=∠ CEB
so
2*∠ AED=360-125*2------> 2*∠ AED=110------> ∠ AED=110/2----> 55°
the answer is
∠ AED= 55°
we know that
The measure of the interior angle is the half-summit of the arcs that comprise it and its opposite.
so
∠ BED=(1/2)*[arc BD+arc CA]-------> (1/2)*[70+180]----> 125°
∠ BED=125°
∠ CEA=∠ BED-----> 125°
∠ AED=∠ CEB
so
2*∠ AED=360-125*2------> 2*∠ AED=110------> ∠ AED=110/2----> 55°
the answer is
∠ AED= 55°
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