Given that, \( \Large tan \theta =\frac{P}{B}=\frac{3}{4} \) and

\( \Large 0 ^{\circ} <0<90 ^{\circ} \) (acute)

In \( \Large \triangle ABC \),

By, Pythagoras theorem

\( \Large BC^{2}=AC^{2}+AB^{2} \)

= 9 + 16 = 25

BC = 5

\( \Large sin \theta =\frac{P}{H}=\frac{3}{5} \)