In the given figure, ∠ BAC = 40°, ∠ ACB = 90° and ∠ BED = 100°. Then ∠ BDE = ?

Question

Philoid · Accepted Answer

In triangle AEF,

BED = EFA + EAF

EFA = 100 – 40 = 60^o

CFD = EFA (vertical opposite angles)

= 60^o

In triangle CFD, we have

CFD + FCD + CDF = 180^o

CDF = 180^o – 90^o – 60^o

= 30^o

So, BDE = 30^o

In the given figure, ∠BAC = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?