1. From an aeroplane just over a straight road, the angles of depression of two consecutive kilometre stones situated at opposite sides of the aeroplane were found to be 60° and 30° respectively. The height (in km) of the aeroplane from the road at that instant was (Given √3 = 1.732)
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AB = 2 and let AC = $$x$$ => BC = $$(2-x)$$ From, $$ riangle$$OAC $$tan60^{circ} = frac{OC}{AC}$$ => $$sqrt{3} = frac{h}{x}$$ => $$x = frac{h}{sqrt{3}}$$ ------------Eqn(1) From, $$ riangle$$OBC $$tan30^{circ} = frac{OC}{BC}$$ => $$frac{1}{sqrt{3}} = frac{h}{2-x}$$ => $$sqrt{3}h = 2 - frac{h}{sqrt{3}}$$ [From eqn(1)] => $$frac{3h+h}{sqrt{3}} = 2$$ => $$h = frac{2sqrt{3}}{4} = frac{sqrt{3}}{2}$$ = $$frac{1.732}{2}$$ = 0.866