Question

in the given figure zBAC= 32°,zBCE= 161°.find zACB,zABC,and zDBC​

Answer 1

Answer:

From the question,

CE∥BA,∠BAC=80o,∠ECD=35o

Now,

(i) ∠BAC=∠ACE=80o    ....[∴ Alternate angles]

(ii) ∠ACB,

=∠ACB+∠ACD=180o     ....[∴ Linear pair]

=∠ACB+∠ACE+∠ECD=180o

=∠ACB+80+35=180

=∠ACB+125=180

=∠ACB=180−115

=∠ACB=65o

(iii) ∠ABC

Let us consider △ABC,

 ∠ABC+∠ACB+∠BAC=180o

=∠ABC+65+80=180o

=∠ABC+145=180

=∠ABC=180−145

=∠ABC=35o