Question
In a $\triangle\text{ABC},$ if a = 18, b = 24, c = 30, find $\cos\text{A},\cos\text{B}$ and $\cos\text{C}.$

Answer

In any $\triangle\text{ABC},$ we have
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{ab}}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
we have,
a = 18, b = 24, c = 30
Therefore,
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}=\frac{1152}{1440}=\frac{4}{5}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}=\frac{648}{1080}=\frac{3}{5}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=\frac{0}{864}=0$

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