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# The angle of elevation of the top of a tree of height 18 meters is $30^\circ$ when measured from a point $P$ in the plane of its base. The distance of the base of the tree from $P$ isA. 6 mB. $6\sqrt 3$mC. 19 mD. $18\sqrt 3$m

Last updated date: 13th Jun 2024
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Hint: The given scenario can be represented by a figure. Draw the diagram of the given problem statement for a better understanding of the situation. Use the formula $\sin \theta = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}$ and $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$ to find the value of the distance between the tree and the point $P$.

Complete step by step solution
Let us assume the distance between the tree and the point $P$ be $x$ meters. It is given in the question that the angle of elevation of the top of the tree from the point $P$ is $30^\circ$. We can therefore draw the diagram representing the scenario.

Here the line TA represents the tree height of 18 meters and the point $P$ is the point in the plane of the tree's base.

We know that in a right angled triangle the $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$

In the triangle$TPA$,
$\tan {30^ \circ } = \dfrac{{TA}}{{PA}}$
Substituting the value 18 for $TA$ and $x$ for $PA$ in the equation, we get
$\tan {30^ \circ } = \dfrac{{18}}{x}$
Solving the above equation for $x$
$\dfrac{1}{{\sqrt 3 }} = \dfrac{{18}}{x} \\ x = 18\sqrt 3 \\$
Thus the horizontal distance of the base of the tree from $P$ is $18\sqrt 3$m.
Therefore the option D is the correct answer.

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 $\theta$.