Question

in the given figure, angle BAC = 32°, Angle BCE = 161°, find angle ACB, Angle ABC and angle DBC

Answer 1

Hey mate!! Here is ur answer
ACB = 19°
ABC = 129°
And
DBC = 51°
I hope it may help u

Answer 2

Answer:

Sol: In ABC , By Exterior Angle Property ;

Exterior angle = sum of opposite interior angles

BCE = ABC + BAC

161 ° = ABC + 32°

161° - 32° = ABC

ABC =129°

Now, By linear pair property of angle ;

ABC + DBC = 180°

129° + DBC = 180°

DBC = 180° - 129°DBC = 51°

Now, By linear pair property of angle;

BCE + BCA = 180°

161° + BCA = 180°

BCA = 180° - 161°

BCA = 19°

please mark it as brainliest answer pls please and give thanks