question_answer

At a point A, 20 m above the level of water in a lake, the angle of elevation of a cloud is \[30{}^\circ \]. The angle of depression of the reflection of the cloud in the lake, at A is \[60{}^\circ \]. Find the distance of the cloud from A.

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 \[40\text{ }m\].