In $\triangle ABC$,right-angled at $B$,$AB = 24 \, cm$,$BC = 7 \, cm$. Determine:
$(i)$ $\sin A, \cos A$
$(ii)$ $\sin C, \cos C$

(N/A) Applying Pythagoras theorem for $\triangle ABC$,we obtain:
$AC^2 = AB^2 + BC^2$
$AC^2 = (24 \, cm)^2 + (7 \, cm)^2$
$AC^2 = (576 + 49) \, cm^2 = 625 \, cm^2$
$\therefore AC = \sqrt{625} \, cm = 25 \, cm$
$(i)$ $\sin A = \frac{\text{Side opposite to } \angle A}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$
$\cos A = \frac{\text{Side adjacent to } \angle A}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$
$(ii)$ $\sin C = \frac{\text{Side opposite to } \angle C}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$
$\cos C = \frac{\text{Side adjacent to } \angle C}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$