The angles of depression of a top and bottom of a building 50 meters high as observed from the top of the tower are 30∘ and 60∘ respectively. Find the height of the tower and also the horizontal distance between the building and the tower respectively.

The angles of depression of a top and bottom of a building 50 meters high as observed from the top of the tower are given as

3 0^{\circ}

and

6 0^{\circ}

respectively.
Let the height of the tower be h and the distance between the building and the tower is x metres. The diagram will be as follows,

We know that for a right-angle triangle, tan

θ

\frac{perpendicular}{base}

.
From the diagram we can write,
tan 60° =

\frac{EC}{BC}

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

\Rightarrow x = \frac{h}{\sqrt{3}}

………. (1) tan 30° =

\frac{ED}{AD}

⇒

\frac{1}{\sqrt{3}}

\frac{h - 50}{x}

x = \sqrt{3} (h - 50)

… (2)
From equation (1) and (2) we get,

\Rightarrow \frac{h}{\sqrt{3}} = \sqrt{3} (h - 50)

\Rightarrow h = 3 (h - 50)

\Rightarrow h = 3 h - 150

\Rightarrow 2 h = 150

\Rightarrow h = 75 m

Putting h = 75 m in (1) we get,

\Rightarrow x = \frac{75}{\sqrt{3}}

\Rightarrow x = 25 \sqrt{3} m

Therefore, height of the tower is 75 m and distance between the tower and the building is

25 \sqrt{3}

m.
Hence, the required option is 1.

The angles of depression of a top and bottom of a building 50 meters high as observed from the top of the tower are $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower and also the horizontal distance between the building and the tower respectively.