Question

Find x° from the following figure​

Answer 1

Answer:

in ∆ abe,

angle bae=54°

angle abe=90°

angle aeb+angle bae+angle abe=180° ( sum of all angles of∆)

angle aeb+54°+90°=180°

angle aeb=180°-144°

=36°

a/q

e is the bisector of angle a

angle a =angle aeb+angle ceb

=36°+36°

=72°

now,in ∆ aec

angle ace+angle a+angle e=180°

angle ace+54°+72°=180°

angle ace=180°-126°

=54°

now,

angle ace+angle dce=180° (by linear pair)

54°+x°=180°

x°=180°-54°

=126°