Question

From the top of a 9 m high building, the angle of elevation of the top of a cable tower is \(60 ^\circ\) and angle of depression of its foot is \(45^\circ\). Determine the height of the tower and distance between building and tower. (use \(\sqrt {3} = 1.732\))

Answer 1

Given, 

Height of building \(=9 \mathrm{~m}\)

Let \(A C=h \mathrm{~m}\) and \(B D=x \mathrm{~m}\)

In \(\triangle B D E,\)

\(\tan 45^{\circ}=\frac{E D}{B D}\)

\(1=\frac{9}{x}\)

x = 9m

In \(\triangle ACE\)

\(tan \,60 ^\circ = \frac {AC}{CE}\)

\(\sqrt {3} = \frac {h}{x}\)

\(h = x\sqrt {3}\)

= 9 x 1.732

= 15.588 m

\(\therefore \) Height of the tower (AB) \(= AC +CB\)

\(= (15.588 + 9) m\)

= 24.588 m.

Answer 2

Let AB be the building and CD be the tower.

Here \(\tan 60^{\circ}=\sqrt{3}=\frac{h}{x}\)

\(\Rightarrow h=x \sqrt{3} \ldots \ldots \) (i)

\(\tan 45^{\circ}=\frac{9}{x}=1\)

\(\Rightarrow x=9 \mathrm{~m} \ldots \ldots \) (ii) (Distance between tower and building)

Solving (i) & (ii) to get h = 9 x 1.732 = 15.588 m

Therefore, the height of the tower = h + 9 = 24.588m.