In Fig., the sides BC, CA and AB of a Δ ABC have been produced to D,

1.	In Fig., 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» Given, ∠ACD = 105° ∠EAF = 45° ∠EAF = ∠BAC (Vertically opposite angle) ∠BAC = 45° ∠ACD + ∠ACB = 180°(Linear pair) 105° + ∠ACB = 180° ∠ACB = 180° – 105° = 75° In ΔABC ∠BAC + ∠ABC + ∠ACB = 180° 45° + ∠ABC + 75° = 180° ∠ABC = 180° – 120° = 60° Thus, all three angles of a triangle are 45°, 60° and 75°

In Fig., 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»

Given,

∠ACD = 105°

∠EAF = 45°

∠EAF = ∠BAC (Vertically opposite angle)

∠BAC = 45°

∠ACD + ∠ACB = 180°(Linear pair)

105° + ∠ACB = 180°

∠ACB = 180° – 105°

= 75°

In ΔABC

∠BAC + ∠ABC + ∠ACB = 180°

45° + ∠ABC + 75° = 180°

∠ABC = 180° – 120°

= 60°

Thus, all three angles of a triangle are 45°, 60° and 75°