yeah.. baby, do the SOH CAH TOA with the hula hoop
[tex]\bf
sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
[/tex]
so.. hmmm we have in the picture, an angle, and two sides,
actually, is the "adjacent" side, and the "opposite"
so.. which of those ratios give us only
the angle
the adjacent side
the opposite side?
ahhha!, is Ms tangent
so [tex]\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(42^o)=\cfrac{h}{100}[/tex]
solve for "h"
