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# The angle of depression of the top and the bottom of a building 50m high are observed from the top of a tower are $30^\circ$ and $60^\circ$ respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Last updated date: 16th May 2024
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Hint: Draw the diagram of the given problem statement for a better understanding of the situation. Use the trigonometric ratios, that are $\sin \theta = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}$ and $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$ in the physical triangle formed to find the height of the tower and horizontal distance between the building and the tower.

Since it is given that the angle observed from the top of the tower to the top and bottom of the building are $30^\circ$ and $60^\circ$respectively, we can draw a diagram representing the condition.

In the diagram, A represents the top of the tower, AB represents the height of the tower, DC represents the building of height 50 metres.
It is given in the question that the angle $\angle TAD$ is $30^\circ$ and the angle $\angle TAC$ is $60^\circ$.
By the property of the corresponding angles of the parallel lines, we can say that the angle $\angle ADE$is $30^\circ$ and the angle $\angle ACB$ is $60^\circ$.
In the triangle $ADE$, we can say
$\tan {30^ \circ } = \dfrac{{{\text{AE}}}}{{DE}}$
On further simplifying
$\dfrac{1}{{\sqrt 3 }} = \dfrac{{{\text{AE}}}}{{DE}} \\ \sqrt 3 {\text{AE = DE}} \\$
Similarly, in the triangle ${\text{ACB}}$
$\tan {60^ \circ } = \dfrac{{{\text{AB}}}}{{{\text{BC}}}}$
On further simplifying
$\sqrt 3 = \dfrac{{{\text{AB}}}}{{{\text{BC}}}} \\ \sqrt 3 {\text{BC = AB}} \\$
From the figure we infer that,
${\text{BC = DE}}$
${\text{AB = AE + EB}} \\ {\text{EB = CD = 50}} \\ {\text{AB = AE + 50}} \\$
Substituting the value ${\text{AE + 50}}$ for ${\text{AB}}$ and ${\text{DE}}$ for ${\text{BC}}$ in the equation $\sqrt 3 {\text{BC = AB}}$, we get
$\sqrt 3 {\text{DE = AE + 50}}$
Also, $\sqrt 3 {\text{AE = DE}}$
Thus the expression becomes $\sqrt 3 \left( {\sqrt 3 {\text{AE}}} \right){\text{ = AE + 50}}$
We can solve the expression to find the value of ${\text{AE}}$
$3{\text{AE = AE + 50}} \\ {\text{2AE = 50}} \\ {\text{AE = 25}} \\$
Substituting the value 25 for ${\text{AE}}$ in the equation ${\text{AB = AE + 50}}$ to find the height of the tower, and in the equation $\sqrt 3 {\text{AE = DE}}$ to find the horizontal distance between the building and the tower.
${\text{AB = 25 + 50}} \\ {\text{AB = 75}} \\$
${\text{DE}} = \sqrt 3 \left( {25} \right)$
Thus the height of the tower is 75 m, and the horizontal distance between the building and the tower is $25\sqrt 3$m.

Note: In a right angled triangle, the $\tan \theta$ is the equal to $\dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$, where perpendicular is the side opposite to the angle $\theta$, and $\sin \theta$ is the equal to $\dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$, where perpendicular is the side opposite to the angle $\theta$.