Question
In $\triangle A B C, \angle A=90^{\circ}$. If $A B=5$ units and $A C=12$ units, find: $\cos C$
✓
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 =1$3 units
$AB =5$ units
$\cos C$
$=\frac{\text { Base }}{ Hypotenuse }$
$=\frac{ AC }{ BC }$
$=\frac{12}{13}$.
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