Answer:
In \[\Delta \text{ }CMP\] \[\tan \,\,30{}^\circ =\frac{CM}{PM}\] \[\frac{1}{\sqrt{3}}=\frac{h}{PM}\] or \[PM=\sqrt{3}h\] ?(i) In \[\Delta \,PMC'\] \[\tan \,\,60{}^\circ =\frac{C'M}{PM}\] \[=\frac{h+60+60}{PM}=\sqrt{3}\] Or \[PM=\frac{h+120}{\sqrt{3}}\] ?(ii) From (i) and (ii) \[\sqrt{3}h=\frac{h+120}{\sqrt{3}}\] \[\Rightarrow \] \[3h=h+120\] \[2h=120\Rightarrow h=60\,\,m\] Height of cloud from surface of water \[=h+60=60+60=120\text{ }m\].
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