In a right angled ΔABC, ∠ABC = 90°, AB = 3, BC = 4, CA = 5; BN is perpendicular to AC, AN : NC is

Solution(By Examveda Team)

According to question,
Given : ∠ABC = 90°
$$\frac{{AN}}{{NC}} = ?$$

ΔABC ∼ ΔBNC
ΔABC ∼ ΔANB
∴ ΔABC ∼ ΔBNC ∼ ΔANB
AB = 3, BC = 4, AC = 5

$$\eqalign{ & \frac{{AB}}{{BN}} = \frac{{AC}}{{BC}} \cr & BN = \frac{{AB \times BC}}{{AC}} = \frac{{3 \times 4}}{5} = 2.4 \cr & \frac{{BC}}{{NC}} = \frac{{AB}}{{NB}} \cr & \frac{4}{{NC}} = \frac{3}{{2.4}} \cr & NC = 3.2 \cr & \frac{{AB}}{{AN}} = \frac{{BC}}{{NB}}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{{AN}} = \frac{4}{{2.4}} \cr & AN = 1.8 \cr & \frac{{AN}}{{NC}} = \frac{{1.8}}{{3.2}} \cr & \frac{{AN}}{{NC}} = \frac{9}{{16}} \cr & \therefore AN:NC = 9:16 \cr} $$