Question

In the figure given below, $ABC$ is an isosceles
triangle with $BC=8$ cm and $AB=AC=5$ cm. Find :

$\sin B$ is $\frac{3}{m}$, m is

Answer 1

$AB=AC=5$, $BC=8$
Using cosine rule,
$A{C}^2=B{C}^2+A{B}^2-2\left(AB\right)\left(BC\right)\cos B$
$5^2=8^2+5^2-2\left(5\right)\left(8\right)\cos B$
$\cos B=\frac{8}{10}$
$\cos B=\frac{4}{5}$
$\cos B=\frac{B}{H}=\frac{4}{5}$
Now, Using Pythagoras Theorem,
$H^2=P^2+B^2$
$5^2=P^2+4^2$
$P=3$
Thus, $\sin B=\frac{3}{5}$