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

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}$
$(ii)$
$\sin C=\frac{\text { Side opposite to } \angle C }{\text { Hypotenuse }}=\frac{A B}{A C}$
$=\frac{24}{25}$
$\cos C=\frac{\text { Side adjacent to } \angle C}{\text { Hypotenuse }}=\frac{B C}{A C}$
$=\frac{7}{25}$