the angle of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are $30^\circ$ and $60^\circ$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

the angle of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are $30^\circ$ and $60^\circ$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $\sqrt{3} = 1.73$)

Let the height of the multi-storeyed building be H  meters.
Let the distance between the two buildings be } d meters.

$ \text{Given:} $
$ \text{Height of the smaller building } = 9 \text{ m} $
Angle of depression to the top of the smaller building = \theta_1
Angle of depression to the foot of the smaller building = \theta_2
$ \text{Using } \sqrt{3} \approx 1.73 $

From the top of the multi-storeyed building, two right-angled triangles are formed.

$ \text{In } \triangle ABC, \tan \theta_1 = \frac{(H - 9)}{d} $
$ \Rightarrow H - 9 = d \tan \theta_1 $
$ \Rightarrow H = 9 + d \tan \theta_1 $

$ \text{In } \triangle ABD, \tan \theta_2 = \frac{H}{d} $
$ \Rightarrow H = d \tan \theta_2 $

$ \text{Equating both equations:} $
$ d \tan \theta_2 = 9 + d \tan \theta_1 $
$ d (\tan \theta_2 - \tan \theta_1) = 9 $
$ d = \frac{9}{\tan \theta_2 - \tan \theta_1} $

$ \text{Now, substituting } d \text{ in } H = d \tan \theta_2, $
$ H = \frac{9 \tan \theta_2}{\tan \theta_2 - \tan \theta_1} $

Thus, the height of the multi-storeyed building is H meters, and the distance between the two buildings is } d meters.

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