NTSE Stage 2 SAT 2016 Question Paper

Draw BL perpendicular to AC and join L to D.
Since ∠BCL=30° we get ∠CBL=60°.
since BLC is right tringle with ∠BCL=30°,we have $\mathrm{BL}=\frac{\mathrm{BC}}{2}=\mathrm{BD}$. Thus in triangle BLD, we observe that BL=BD and ∠DBL=60° and ∠ADB=45°, we get ∠ADL=15°.
But ∠DAL=15°.Thus LD=LA.we hence have LD=LA=LB. This implies that L is the circumcenter of the triangle BDA.
Thus $\angle \mathrm{BAD}=\frac{1}{2}\angle \mathrm{BLD}=\frac{1}{2}\times 60°=30°$
$30°+45°+\angle \mathrm{ABC}=180°$
Hence ∠ABC=105°
so the value of ∠BAD and ∠ABC are 30° and 105° respectively