A helicopter is flying at an altitude of $250m$ between two banks of a river. From the helicopter it is observed that the angles of a river. From the helicopter its is observed  that the angle of depression of two boats on the opposite banks are $45^o$ and $60^o$ respectively. Find the width of the river. Given your answer correct to nearest metre.

#### Solution

Now,
$\tan { { 30 }^{ \circ } } =\cfrac { HC }{ AC } =\cfrac { 250 }{ AC } \quad ;\quad \quad \tan { { 45 }^{ \circ } } =\cfrac { HC }{ BC } \\ \therefore AC=250\sqrt { 3 } m\quad \quad ;HC=BC=250m$
$\therefore$ Width of the river is $=\left| AB \right| =250\sqrt { 3 } +250$
$=250(\sqrt { 3 } +1)\\ \cong 683m$