Question

In triangle $ABC$, $AB=AC=15$ cm and $BC=18$ cm. Find :

m if $\sin C$ is $\frac{4}{m}$,

Answer 1

$AB=AC=15$, $BC=18$
Using cosine rule,
$A{B}^2=B{C}^2+A{C}^2-2\left(AB\right)\left(BC\right)\cos C$
${15}^2={18}^2+{15}^2-2\left(15\right)\left(18\right)\cos C$
$\cos C=\frac{9}{15}$
$\cos C=\frac{3}{5}$
$\cos C=\frac{B}{H}=\frac{3}{5}$
Using Pythagoras Theorem,
$H^2=P^2+B^2$
$5^2=P^2+3^2$
$P=4$
$\sin C=\frac{P}{H}=\frac{4}{5}$