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.

Answer 1

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°.