Question

If sin A = $\frac{9}{41}$, find the values of cos A and tan A,

Answer 1

Consider the right-angled triangle ABC,

Let x be the adjacent side
By applying Pythagorean

Given $sinA=\frac{9}{41}$

$sinA=\frac{Oppositeside}{Hypotenuseside}=\frac{9}{41}$

Let x be the adjacent side
By applying Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

${41}^2=x^2+9^2$

$x^2={41}^2-9^2$

$x^2=1681-81=1600$

$x=\sqrt{1600}=40$

$cosA=\frac{Adjacentside}{Hypotenuseside}=\frac{40}{41}$
$tanA=\frac{Oppositeside}{Adjacentside}=\frac{9}{40}$