Respuesta :

notice the picture below

you have two triangles, and you have the opposite and adjacent sides

thus, recall your SOH CAH TOA   [tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\\\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}[/tex]

anyhow.. the one that has only
the angle
the opposite
and the adjacent

is Ms Tangent, so lets ask her some

[tex]\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)[/tex]

solve for "y"
Ver imagen jdoe0001

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