Question
In $\triangle ABC, \angle A = 90^\circ.$ If $AB = 5$ units and $AC = 12$ units, find$: \sin B$
✓
Answer
In $\triangle ABC$
$BC ^2= AB ^2+ AC ^2$
$\Rightarrow BC =\sqrt{ AB ^2+ AC ^2}$
$\Rightarrow BC =\sqrt{5^2+12^2}$
$=\sqrt{169}$
$=13$
$AC =12$ units
$BC =13$ units
$AB =5$ units
$\sin B$
$ =\frac{\text { Perpndicular }}{\text { Hypoenuse }}$
$=\frac{ AC }{ BC }$
$=\frac{12}{13}$
and
$\cos B$
$=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{ AB }{ BC }$
$=\frac{5}{13} . $
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