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A balloon is moving up in air vertically above a point $A$ on the ground. When it is at a height $h _{1},$ a girl standing at a distance $d$ (point $B$ ) from $A$ (see figure) sees it at an angle $45^{\circ}$ with respect to the vertical. When the balloon climbs up a further height $h _{2},$ it is seen at an angle $60^{\circ}$ with respect to the vertical if the girl moves further by a distance $2.464\, d$ (point $C$ ). Then the height $h _{2}$ is (given tan $\left.30^{\circ}=0.5774\right)$$.......$
$d$
$0.732d$
$1.464d$
$0.464d$
Solution
$\frac{ h _{1}}{ d }=\tan 45^{\circ} \Rightarrow h _{1}= d \ldots(1)$
$\frac{ h _{1}+ h _{2}}{ d +2.464 d }=\tan 30^{\circ}$
$\Rightarrow\left( h _{1}+ h _{2}\right) \times \sqrt{3}=3.46 d$
$\left(h_{1}+h_{2}\right)=\frac{3.46 d }{\sqrt{3}}$
$\Rightarrow d + h _{2}=\frac{3.46 d }{\sqrt{3}}$
$h _{2}= d$
