Question

in the figure ab parallel to de. find angle ABC=75 and angle cde =145, then find angle bcd

Answer 1

angle BCD =75-(180-145)
=75-35
=45

Answer 2

Answer:

∠BCD =40°

Step-by-step explanation:

Refer the attached figure

Extend the line AB to F

So, AF || DE

DC is traversal

∠EDC=∠BOD (Alternate interior angles)

∠EDC=∠BOD = 145°

∠BOD+∠BOC =180° (Linear pair)

145°+∠BOC =180°

∠BOC =180°-145°

∠BOC =35°

In BOC

∠BOC+∠BCO = ∠ABC (exterior angle property of triangle)

35°+∠BCO = 75°

∠BCO = 75°-35°

∠BCO =40°

So, ∠BCD =40°