is a triangle in which AB = AC . Let BC be produced to D . From a point E on the line AC let EF be a straight line such that EF is parallel to AB . Consider the quadrilateral ECDF thus formed. If ∠ABC = 65∘ and ∠EFD = 80∘ , then what is ∠FDC equal to?
Question
is a triangle in which AB = AC . Let BC be produced to D . From a point E on the line AC let EF be a straight line such that EF is parallel to AB . Consider the quadrilateral ECDF thus formed. If ∠ABC = 65∘ and ∠EFD = 80∘ , then what is ∠FDC equal to?
Solution
Given that AB = AC, we have an isosceles triangle ABC.
Let BC be produced to point D.
From a point E on the line AC, draw a straight line EF parallel to AB.
This forms the quadrilateral ECDF.
We are given that ∠ABC = 65° and ∠EFD = 80°.
To find ∠FDC, we can use the fact that the sum of the angles in a quadrilateral is 360°.
Since ∠ABC and ∠EFD are already known, we can find ∠FDC by subtracting the sum of the other two angles from 360°.
∠FDC = 360° - ∠ABC - ∠EFD
∠FDC = 360° - 65° - 80°
∠FDC = 215°
Therefore, ∠FDC is equal to 215°.
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