Question

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

$\tan C-\cot B$ is equal to $\frac{-7}{m}$, then $m$ 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, $\cot B=\frac{B}{P}=\frac{4}{3}$
Now, to find $\tan C$
Using cosine rule,
$A{B}^2=A{C}^2+B{C}^2-2\left(AC\right)\left(BC\right)\cos C$
$5^2=5^2+8^2-2\left(5\right)\left(8\right)\cos C$
$\cos C=\frac{8}{10}$
$\cos C=\frac{4}{5}$
$\cos C=\frac{B}{H}=\frac{4}{5}$
Now, Using Pythagoras Theorem,
$H^2=P^2+B^2$
$5^2=P^2+4^2$
$P=3$
Thus, $\tan C=\frac{P}{B}=\frac{3}{4}$
Thus. $\tan C-\cot B=\frac{3}{4}-\frac{4}{3}$
= $\frac{9-16}{12}$
= $-\frac{7}{12}$