In traveling across flat land, you see a mountain directly in front of you. Its angle of elevation (to the peak) is 3.5°. After you drive 15 miles closer to the mountain, the angle of elevation is 9° (see figure). Approximate the height of the mountain. (Round your answer to one decimal place.)

check the picture below.

[tex]\bf tan(3.5^o)=\cfrac{y}{15+x}\implies 15+x=\cfrac{y}{tan(3.5^o)}\\\\\\ \boxed{x=\cfrac{y}{tan(3.5^o)}-15} \\\\\\ tan(9^o)=\cfrac{y}{x}\implies x\cdot tan(9^o)=y\\\\\\ \left[ \cfrac{y}{tan(3.5^o)}-15 \right]tan(9^o)=y \\\\\\ \left[ \cfrac{y-15tan(3.5^o)}{tan(3.5^o)} \right]tan(9^o)=y \\\\\\[/tex]

[tex]\bf \cfrac{y\cdot tan(9^o)-15tan(3.5^o)tan(9^o)}{tan(3.5^o)}=y \\\\\\ y\cdot tan(9^o)-15tan(3.5^o)tan(9^o)=y\cdot tan(3.5^o) \\\\\\ ytan(9^o)-ytan(3.5^o)=15tan(3.5^o)tan(9^o) \\\\\\ \stackrel{common~factor}{y}[tan(9^o)-tan(3.5^o)]=15tan(3.5^o)tan(9^o) \\\\\\ y=\cfrac{15tan(3.5^o)tan(9^o)}{tan(9^o)-tan(3.5^o)}[/tex]

make sure your calculator is in Degree mode.

the amount is small, but bear in mind that is mile units.