Question

In triangle $ABC$, $AB=AC=15$ cm and $BC=18$ cm, find $\cos \angle ABC$ is $\frac{3}{m}$, m is

Answer 1

$AB=AC=15$, $BC=18$
Using cosine rule,
$A{C}^2=B{C}^2+A{B}^2-2\left(AB\right)\left(BC\right)\cos B$
${15}^2={18}^2+{15}^2-2\left(15\right)\left(18\right)\cos B$
$\cos B=\frac{9}{15}$
$\cos B=\frac{3}{5}$