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Question

In the given figure, ABC is an isosceles triangle in which AB = AC. AEDC is a parallelogram. If ∠CDF = 70° and ∠BFE = 100°, then ∠FBA = ____________.

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Solution

Given:
ABC is an isosceles triangle
AB = AC
AEDC is a parallelogram
∠CDF = 70°
∠BFE = 100°


AEDC is a parallelogram
∠ACD + ∠CDE = 180° (interior angles)
⇒ ∠ACD + 70° = 180°
⇒ ∠ACD = 180° − 70°
⇒ ∠ACD = 110° ....(1)


Now, ∠ACD + ∠ACB = 180° (angles on a straight line)
⇒ 110° + ∠ACB = 180°
⇒ ∠ACB = 180° − 110°
⇒ ∠ACB = 70° ....(2)

Also, ∠BFE + ∠BFD = 180° (angles on a straight line)
⇒ 100° + ∠BFD = 180°
⇒ ∠BFD = 180° − 100°
⇒ ∠BFD = 80° ....(3)


Now, in ∆BFD
∠FBD + ∠BDF + BFD = 180° (angle sum property)
⇒ ∠FBD + 70° + 80° = 180°
⇒ ∠FBD + 150° = 180°
⇒ ∠FBD = 180° − 150°
⇒ ∠FBD = 30° ....(4)

Since, ABC is an isosceles triangle with AB = AC
Thus, ∠ABC = ∠ACB = 70° (From (2))

∠ABC = ∠ABF + ∠FBD
⇒ 70° = ∠ABF + 30°
⇒ ∠ABF = 70° − 30°
⇒ ∠ABF = 40°


Hence, ∠FBA = 40°.

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