If the angle of elevation of a cloud from a point 100 m above a lake is 45° and the angle of depression of its reflection in the lake is 60°. Then the height of the cloud above the lake is?
Answer & explanation
Correct answer: option 3
Here, BD = 100m, let, AD = (x) m, DC = 100 + x
In Δ PDC;
tan 60° = \(\frac{DC}{PD}\)
⇒ \(\frac{\sqrt {3}}{1}\) = \(\frac{100 + x}{PD}\) .......(i)
and
In Δ PDA;
tan 45° = \(\frac{AD}{PD}\)
⇒ \(\frac{1}{1}\) = \(\frac{x}{PD}\) .........(ii)
Dividing (i) by (ii), we get,
⇒ \(\sqrt {3}\) = \(\frac{100 + x}{x}\)
⇒ x (\(\sqrt {3}\) - 1) = 100
⇒ x = \(\frac{100}{\sqrt {3} - 1}\) = 50 (\(\sqrt {3}\) + 1)
Height of cloud (AB) = AD + DB = (x + 100) = 50 (\(\sqrt {3}\) + 1) + 100
= 50 (\(\sqrt {3}\) + 3) m