Question

6)In the figure below, it is given that ZA = 60°CEIBA and 2 ECD-65° then LACB 65 a)55° (b)90° (c) 60° (d) 70°​

Answer 1

Answer:

AngleBAC=angleACE....(alt interior angle)

AngleBAC=angleACE....(alt interior angle)So,

AngleBAC=angleACE....(alt interior angle)So, Angle ACE=65 dgree

AngleBAC=angleACE....(alt interior angle)So, Angle ACE=65 dgreeAngle ACB+ angleACE + angle ECD= 180 degree....(linear pair)

AngleBAC=angleACE....(alt interior angle)So, Angle ACE=65 dgreeAngle ACB+ angleACE + angle ECD= 180 degree....(linear pair)Angle ACB+65 degree + 60 degree=180 dgree

AngleBAC=angleACE....(alt interior angle)So, Angle ACE=65 dgreeAngle ACB+ angleACE + angle ECD= 180 degree....(linear pair)Angle ACB+65 degree + 60 degree=180 dgreeAngle BCA=(180-125) degree

AngleBAC=angleACE....(alt interior angle)So, Angle ACE=65 dgreeAngle ACB+ angleACE + angle ECD= 180 degree....(linear pair)Angle ACB+65 degree + 60 degree=180 dgreeAngle BCA=(180-125) degreeAngle BCA=55 degree (Ans).