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

Answer: Option B

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

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

$$\begin{array}{rl}& \frac{AB}{BN}=\frac{AC}{BC}\\ & BN=\frac{AB\times BC}{AC}=\frac{3\times 4}{5}=2.4\\ & \frac{BC}{NC}=\frac{AB}{NB}\\ & \frac{4}{NC}=\frac{3}{2.4}\\ & NC=3.2\\ & \frac{AB}{AN}=\frac{BC}{NB}\frac{3}{AN}=\frac{4}{2.4}\\ & AN=1.8\\ & \frac{AN}{NC}=\frac{1.8}{3.2}\\ & \frac{AN}{NC}=\frac{9}{16}\\ & \therefore AN:NC=9:16\end{array}$$