4. Problems involving angle of elevation and depression

Kumar takes walks every evening. One evening, he observes a bird flying with an angle of elevation of \(60^{\circ}\). He also observes a dog right beneath the bird with an angle of depression of \(30^{\circ}\). If the height of Kumar is \(1.5 \ m\), then find the altitude of the bird from the dog.

 

Solution:

 

 

Let \(AB\) denote the height of Kumar. Then, \(AB = 1.5 \ m\).

 

\(CD\) is the altitude of the bird from the ground. Through \(A\), draw \(AO \perp CD\).

 

The angle of elevation is \(\angle OAD = 60^{\circ}\).

 

The angle of depression is \(\angle OAC = 30^{\circ}\).

 

Let \(h\) denote the altitude of the bird from the ground.

 

That is, \(CD = h\).

 

Then, \(OD = CD - OC = h - 1.5\)

 

In right triangle \(OAC\), \(tan \ \theta = \frac{OC}{AO}\)

 

\(tan \ 30^{\circ} = \frac{1.5}{AO}\)

 

\(\frac{1}{\sqrt{3}} = \frac{1.5}{AO}\)

 

\(AO = 1.5\sqrt{3}\) ---- (\(1\))

 

In right triangle \(OAD\), \(tan \ \theta = \frac{OD}{AO}\)

 

\(tan \ 60^{\circ} = \frac{h - 1.5}{1.5\sqrt{3}}\) [Using equation (\(1\))]

 

\(\sqrt{3} = \frac{h - 1.5}{1.5\sqrt{3}}\)

 

\(1.5\sqrt{3}(\sqrt{3}) = h - 1.5\)

 

\(1.5 \times 3 = h - 1.5\)

 

\(4.5 = h - 1.5\)

 

\(4.5 + 1.5 = h\)

 

\(6 = h\)

 

Therefore, the altitude of the bird from the dog is \(6 \ m\).