Mathematics

In the above figure, lines $$AB$$ and $$CD$$ intersect at $$F$$. If $$\angle EFA=\angle AFD$$ and $$\angle CFB=50^{0}$$, find $$\angle EFC$$.


SOLUTION
In the given figure below,
$$AB$$ and $$CD$$ intersect each other at $$F$$.
$$\angle EFA=\angle AFD$$ and $$\angle CFB=50^{0}$$
To find: $$\angle EFC$$
$$\angle AFD=\angle BFC=50^{0}$$   (Vertically opposite angles)
But, $$\angle EFA=\angle AFD$$ (given) = $$50^{0}$$.
Now, $$\angle EFA+\angle EFC+\angle CFB=180^{0}$$
(Angles on one side of straight line)
$$\Rightarrow 50^{0}+\angle EFC+50^{0}=180^{0}$$
$$\Rightarrow \angle EFC+100^{0}=180^{0}$$
$$\Rightarrow \angle EFC=180^{0}-100^{0}=80^{0}$$
$$\Rightarrow \angle EFC=80^{0}$$.
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Subjective Medium Published on 09th 09, 2020
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