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Question
In the given figure, the sides BC, CA and AB of a Δ ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the Δ ABC.
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Solution
In the given ΔABC, ∠ACD = 105°and ∠EAF = 45°. We need to find ∠ABC, ∠ACB, and ∠BAC.
Here, ∠EAF and ∠BAC are vertically opposite angles. So, using the property, “vertically opposite angles are equal”, we get,
∠EAF = ∠BAC
∠BAC = 45°
Further, BCD is a straight line. So, using linear pair property, we get,
∠ACB + ∠ACD = 180°
∠ACB + 105° = 180°
∠ACB = 180° - 105°
∠ACB = 75°
Now, in ΔABC, using “the angle sum property”, we get,
∠ABC + ∠ACB + ∠BAC = 180°
45° + 75° + ∠ABC = 180°
∠BAC = 180°
∠BAC = 180° - 120°
∠BAC = 60
Therefore,∠ACB = 75° , ∠BAC = 45°,∠ABC = 60°.
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