Question

in figure ab is equal to CD is equal to BC and BC is equal to 96 degree then find angle DBC

Answer 1

Let ∠DAC = ∠ACD = α and ∠CDB = ∠CBD = β.

As ∠CDB is the exterior angle of triangle ACD,
β = 2α.

Now ∠ACD + ∠DCB + 96° = 180°
⇒ α + 180° − 2β + 96° = 180°
⇒ 3α = 96°
⇒ α = 32°
⇒ β = 64°

Hope This Helps :)