Question

Refer to the triangle given in figure.

$\sec C$ is $\frac{3\sqrt{2}}{m}$, m is

Answer 1

Given, $△ABC$, A perpendicular from B on AC, let it cut AC at D, such that$\angle BDA=\angle BDC={90}^{\circ }$
$AD=3$
$BD=4$
$BC=12$
In $△ABD$,
$A{B}^2=B{D}^2+A{D}^2$
$A{B}^2=3^2+4^2$
$AB=5$
In $△BCD$
$B{C}^2=C{D}^2+B{D}^2$
${12}^2=C{D}^2+4^2$
$C{D}^2=128$
$CD=8\sqrt{2}$
Now, $\sec C=\frac{H}{P}=\frac{12}{8\sqrt{2}}=\frac{3}{2\sqrt{2}}$