Question
In right angled $\triangle \mathrm{ACB}$, If $\angle \mathrm{C}=90^{\circ}, \mathrm{AC}=3, \mathrm{BC}=4$.Find the ratios $\sin A, \sin B, \cos A, \tan B$
✓
Answer
In right angled $\triangle \mathrm{ACB}$, using Pythagoras' theorem,
$\mathrm{AB}^2=\mathrm{AC}^2+\mathrm{BC}^2$
$=3^2+4^2=5^2$
$\therefore \mathrm{AB}=5$
$\sin \mathrm{A}=\frac{B C}{A B}=\frac{4}{5}$
$and \sin \mathrm{B}=\frac{A C}{A B}=\frac{3}{5}$
$\cos \mathrm{A}=\frac{A C}{A B}=\frac{3}{5}$
$\tan \mathrm{B}=\frac{A C}{B C}=\frac{3}{4}$
$\mathrm{AB}^2=\mathrm{AC}^2+\mathrm{BC}^2$
$=3^2+4^2=5^2$
$\therefore \mathrm{AB}=5$
$\sin \mathrm{A}=\frac{B C}{A B}=\frac{4}{5}$
$and \sin \mathrm{B}=\frac{A C}{A B}=\frac{3}{5}$
$\cos \mathrm{A}=\frac{A C}{A B}=\frac{3}{5}$
$\tan \mathrm{B}=\frac{A C}{B C}=\frac{3}{4}$
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