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