Question

In $\Delta ABC,$ right angled at $B$, $AC+AB=9$ cm and $BC=3$ cm.The value of $\cot C=\frac{3}{k}$, $k$ is

Answer 1

In $\Delta ABC$ we have
$(AC{)}^2=(AB{)}^2+B{C}^2$
$\Rightarrow (9-AB{)}^2=A{B}^2+(3{)}^2$
$[\because AC+AB=9cm\Rightarrow AC=9-AB]$
$\Rightarrow 81+A{B}^2-18AB=A{B}^2+9$
$\Rightarrow 72-18AB=0$
$\Rightarrow AB=\frac{72}{18}=4cm$
Now, $AC+AB=9$
$⟹AC=9-4=5$
$\therefore \cot C=\frac{BC}{AB}=\frac{3}{4}$

Comparing it with the given value we get, $k=4$