notice the picture below
you have two triangles, and you have the opposite and adjacent sides
thus, recall your SOH CAH TOA \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\ \quad \\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\)
anyhow.. the one that has only
the angle
the opposite
and the adjacent
is Ms Tangent, so lets ask her some
\(\bf tan(29^o)=\cfrac{x}{y}\implies y\cdot tan(29^o)=x
\\\\\\
tan(25^o)=\cfrac{x}{100-y}\implies (100-y)tan(25^o)=x\\\\
-----------------------------\\\\
x=x\qquad thus\qquad y\cdot tan(29^o)=(100-y)tan(25^o)\)
solve for "y"
