Question

If $\sin A$= $\frac{9}{41},$ compute $\cos A$ and $\tan A$.

Answer 1

Given,

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

We know that,

$sinA=\frac{oppositeSide}{Hypotenuse}$

From Pythagoras theorem,

$(Hypotenuse{)}^2=(oppositeSide{)}^2+(adjacentSide{)}^2$

${41}^2=9^2+(adjacentSide{)}^2$

$(adjacentSide{)}^2=1681-81=1600$

$\left(adjacentSide\right)=40$

$cosA=\frac{AdjacentSide}{Hypotenuse}=\frac{40}{41}$

$tanA=\frac{OppositeSide}{AdjacentSide}=\frac{9}{40}$