In $\triangle ABC ,$ right-angled at $B , AB =24 \,cm , BC =7 \,cm .$ Determine:

$(i)$ $\sin A, \cos A$

$(ii)$ $\sin C, \cos C$

Applying Pythagoras theorem for $\triangle ABC ,$ we obtain

$A C^{2}=A B^{2}+B C^{2}$

$=(24\, cm )^{2}+(7\, cm )^{2}$

$=(576+49) \,cm ^{2}$

$=625\, cm ^{2}$

$\therefore A C=\sqrt{625} cm =25\, cm$

$(i)\,\sin A\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AC }$

$=\frac{7}{25}$

$\cos A=\frac{\text { Side adjacent to } \angle A }{\text { Hypotenuse }}=\frac{ AB }{ AC}$$=\frac{24}{25}$

$(ii)$

$\sin C=\frac{\text { Side opposite to } \angle C }{\text { Hypotenuse }}=\frac{A B}{A C}$

$=\frac{24}{25}$

$\cos C=\frac{\text { Side adjacent to } \angle C}{\text { Hypotenuse }}=\frac{B C}{A C}$

$=\frac{7}{25}$