While flying at an altitude of 1.5 km, a plane measures angles or depression to opposite ends of a large crater, shown in the image below. Find the width of the crater

Accepted Solution

Check the picture below.notice the alternate interior angles in the picture.[tex]\bf tan(68^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{x}}\implies x=\cfrac{1.5}{tan(68^o)}\implies x\approx 0.61 \\\\[-0.35em] ~\dotfill\\\\ tan(56^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{w}}\implies w=\cfrac{1.5}{tan(56^o)}\implies w\approx 1.01 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{width of the crater}}{x+w\implies 1.62}~\hfill[/tex]