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 $$=\left| AB \right| =250\sqrt { 3 } +250$$
$$ =250(\sqrt { 3 } +1)\\ \cong 683m$$