Question

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

Answer 1

$sinA=\frac{9}{41}=\frac{oppositeside}{hypotenuse}(hypotenuse{)}^2=(base{)}^2+(perpendicular{)}^2{41}^2=9^2+(perpendicular{)}^{2}1681-81=1600\sqrt{1600}=40cosA=\frac{adjacentside}{hypotenuse}=\frac{40}{41}tanA=\frac{oppositeside}{adjacentside}=\frac{9}{40}$