UPSC 2016 CDS I Math Paper

In ΔBCD,CD=BC
∴ ∠CBD = ∠CDB....(1) (In a triangle, equal sides have equal angles opposite to them)
In ΔACD , CD = AD
∴ ∠CAD = ∠ACD....(2) (In a triangle, equal sides have equal angles opposite to them)
In ΔABC,
∠CAD + ∠CBD = 96°(Exterior angle of a triangle is equal to the sum of opposite interior angles)
⇒

180°−∠ADC

2

+∠CDB=96°
[Using (1), (2) and angle sum property]
⇒

3

2

∠CDB=96°⇒∠CDB=64°