In the adjoining figure, it is given that CE ∥ BA, ∠BAC = 80°

1.	In the adjoining figure, it is given that CE ∥ BA, ∠BAC = 80° and ∠ECD = 35°.Find (i) ∠ACE, (ii) ∠ACB, (iii) ∠ABC.
Answer» From the question, CE∥BA, ∠BAC = 80°, ∠ECD = 35°. Now, (i) ∠BAC = ∠ACE = 80° … [∵ Alternate angles] (ii) ∠ACB, = ∠ACB + ∠ACD = 180° … [∵ Linear pair] = ∠ACB + ∠ACE + ∠ECD = 180° = ∠ACB + 80 + 35 = 180 = ∠ACB + 125 = 180 = ∠ACB = 180 – 115 = ∠ACB = 65° (iii) ∠ABC Let us consider Δ ABC, ∠ABC + ∠ ACB + ∠BAC = 180° = ∠ABC + 65 + 80 = 180 = ∠ABC + 145 = 180 = ∠ABC = 180 – 145 = ∠ABC = 35°

In the adjoining figure, it is given that CE ∥ BA, ∠BAC = 80° and ∠ECD = 35°.Find (i) ∠ACE, (ii) ∠ACB, (iii) ∠ABC.

Answer»

From the question,

CE∥BA, ∠BAC = 80°, ∠ECD = 35°.

Now,

(i) ∠BAC = ∠ACE = 80° … [∵ Alternate angles]

(ii) ∠ACB,

= ∠ACB + ∠ACD = 180° … [∵ Linear pair]

= ∠ACB + ∠ACE + ∠ECD = 180°

= ∠ACB + 80 + 35 = 180

= ∠ACB + 125 = 180

= ∠ACB = 180 – 115

= ∠ACB = 65°

(iii) ∠ABC

Let us consider Δ ABC,

∠ABC + ∠ ACB + ∠BAC = 180°

= ∠ABC + 65 + 80 = 180

= ∠ABC + 145 = 180

= ∠ABC = 180 – 145

= ∠ABC = 35°