Answer:
Let PQ be the surface of the lake. A is the point vertically above P such that \[AP=20\text{ }m\]. Let C be the position of the cloud and D be its reflection in the lake. Let \[BC=H\]metres Now, In \[\Delta \,ABD\] \[\tan \,60{}^\circ =\frac{BD}{AB}\] \[\Rightarrow \sqrt{3}=\frac{H+20+20}{AB}\] \[\Rightarrow \sqrt{3}.\,\,AB=H+40\] \[\Rightarrow AB=\frac{H+40}{\sqrt{3}}\] ?(i) And, in \[\Delta \text{ }ABC\] \[\tan 30{}^\circ =\frac{BC}{AB}\] \[\frac{1}{\sqrt{3}}=\frac{H}{AB}\] \[AB=\sqrt{3}H\] ?(ii) From eq. (i) and (ii) \[\frac{H+40}{\sqrt{3}}=\sqrt{3}H\] \[\Rightarrow 3H=H+40\] \[\Rightarrow 2H=40\Rightarrow H=20\] Putting the value of H in eq. (ii), we get \[AB=20\sqrt{3}\] Again, in \[\Delta \text{ }ABC\] \[{{(AC)}^{2}}={{(AB)}^{2}}+{{(BC)}^{2}}\] \[={{\left( 20\sqrt{3} \right)}^{2}}+{{(20)}^{2}}\] \[=1200+400\] \[=1600\] \[AC=\sqrt{1600}=40\] Hence, the distance of cloud from A is \[40\text{ }m\].
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