Question

Find the values of the trigonometric ratio
$\cos A=\frac{4}{5}$

Answer 1

According to question,

$cosA=\frac{4}{5}=\frac{B}{H}$

Let hypotenuse , $H=5k$

So, base,$B=4k$

$P^2+B^2=H^2$

$P^2=H^2-B^2$

$=(5k{)}^2-(4k{)}^2$

$=9k^2$

$P=3k$

$sinA=\frac{P}{H}$
$=\frac{3}{5}$

$cotA=\frac{B}{P}$
$=\frac{4}{3}$

$cosecA=\frac{H}{P}$
$=\frac{5}{3}$

$secA=\frac{H}{B}$
$=\frac{5}{4}$

$tanA=\frac{P}{B}$

$=\frac{3}{4}$