Two poles of equal heights are standing opposite each other on either side of the roads, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are ${60^ \circ }$and ${30^ \circ }$ respectively. Find the height of the poles and the distances of the point from the poles.
VerifiedHint: First of all, create the figure as per the instructions and then start working. Now, assume the lengths required and use the tangent of the angles formula and equate them as per their values and thus we will have our result.
Complete step-by-step answer:
Let us firstly draw the image as per the given required. Let AB and CD be the two poles of equal length with 80 m distance between them that is BD = 80 m. The angles are as mentioned below in the figure:
Now, we need to find the height of the poles that is the length of AB or CD (because they are both equal).
Let AB and CD be equal to h meters.
Let $BO = x$ m. so that $OD = (80 - x)$ m.
Consider $\vartriangle ABO$:
We know that $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$.
So, $\tan {30^ \circ } = \dfrac{{AB}}{{OB}}$
Putting the values AB = h meters and $BO = x$ meters, we will have:-
$\dfrac{1}{{\sqrt 3 }} = \dfrac{h}{x}$
$ \Rightarrow h = \dfrac{x}{{\sqrt 3 }}$ ……….(1)
Consider $\vartriangle CDO$:
We know that $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$.
So, $\tan {60^ \circ } = \dfrac{{CD}}{{OD}}$
Putting the values CD = h meters and $OD = 80 - x$ meters, we will have:-
$\sqrt 3 = \dfrac{h}{{80 - x}}$
$ \Rightarrow h = \sqrt 3 (80 - x)$ ……….(2)
$ \Rightarrow \sqrt 3 (80 - x) = \dfrac{x}{{\sqrt 3 }}$
Rewriting it as follows:-
$ \Rightarrow 3(80 - x) - x = 0$
Combining the like terms to get:-
$ \Rightarrow 240 - 4x = 0$
$ \Rightarrow 4x = 240$
Taking the 4 from LHS to RHS:
$ \Rightarrow x = 60$.
Putting this value in (1), we will get:-
$ \Rightarrow h = \dfrac{{60}}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = 20\sqrt 3 m$
Hence, the heights of the pole is $20\sqrt 3 m$ and the distance between the point of elevation and the poles is 60 m each.
Note: The "upwards" angle from the horizontal to a line of sight from the observer to some point of interest. If the angle goes "downwards" it is called an Angle of Depression.
The students might think that O will be the midpoint of these types of questions always but that is not necessarily true because here, we got that result, if the angles change or the heights of pole change, this may change as well.
The students must note that we can use the trigonometric identities only on right angles, not any random triangle.
