Question

In $△$PQR, PQ = 12 cm, PR = 13 cm and $\angle Q={90}^{\circ }$. Find the value of 12tan P and cot P.

Answer 1

Given that in $△$ PQR, PQ = 12 cm and PR = 13 cm.

Now, from Pythagoras theorem,
${PQ}^2+{QR}^2={PR}^2$
$\Rightarrow {QR}^2={PR}^2-{PQ}^2$
$\Rightarrow {QR}^2={13}^2-{12}^2$
$\Rightarrow {QR}^2=169-144=25$
$\Rightarrow QR=\sqrt{25}=5\text{cm}$

Now, $tanP=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{QR}{PQ}=\frac{5}{12}$

$\therefore 12\tan P=12\times \frac{5}{12}=5$

We know that $\cot \theta$ is the reciprocal of $\tan \theta$, therefore cot P is given as,

$\cot P=\frac{1}{\tan P}=\frac{12}{5}$.