Question

From the following figure, find the value of m, when it is given that $\sec A$ is $\sqrt{m}$.

Answer 1

Given, $△ABC$, $\angle ABC={90}^{\circ }$, $AB=BC=a$
By Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=a^2+a^2$
$AC=a\sqrt{2}$
Now, $\sec A=\frac{H}{P}=\frac{AC}{BC}=\frac{a\sqrt{2}}{a}=\sqrt{2}$