From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and the angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.

Let DE be the surface of lake and A be the point 36 m above surface of lake and F is the image of bird (B).

Let BC be h meters.

BD = DF (As distance of bird from lake = distance of image of bird in lake from lake)

In △ABC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{BC}{AC} \\[1em] \Rightarrow AC = \sqrt{3}BC \\[1em] \Rightarrow AC = \sqrt{3}h.$

From figure,

⇒ CF = CD + DF

⇒ CF = CD + BD

⇒ CF = CD + BC + CD

⇒ CF = 36 + h + 36

⇒ CF = h + 72.

In △ACF,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CF}{AC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h + 72}{\sqrt{3}h} \\[1em] \Rightarrow 3h = h + 72 \\[1em] \Rightarrow 2h = 72 \\[1em] \Rightarrow h = 36 \text{ m}.$

Height of bird from lake = BD = BC + CD = h + 36 = 36 + 36 = 72 m.

Hence, height of bird above surface of lake = 72 meters.