Question

Two poles of equal height stand vertically opposite to each other on either side of the road, which is 100 m wide. From a point on the road between the poles, the angle of elevation of the tops of the poles are 30 degree and 60 degree . Find the heights of the poles. Also find the distance of the point from the feets to the poles.​

Answer 1

Answer:

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$\LARGE\bold{Given :}$

Bottom distance = 100 m.

Angle of Elevation$\bf{\angle_{1}}$

Angle of Elevation$\bf{\angle_{2}}$

$\LARGE\bold{To\: find :}$

Height of the two poles.

Distance of the points from the feet of the poles.

$\LARGE\bold{Solution :}$

Let the height of both the poles will be h m.

Let the distance from point A and B be x m.

Hence according to the Question , the distance from point B will be (100 - x) m.

$\LARGE\bold{Height \:of\: the\: tower :)}$

To find the height of pole (in terms of h) with respect to angle 60°.

Using tan θ and substituting the values in it, we get :

$\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{tan\:60^{\circ} = \dfrac{h}{x}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\sqrt{3} = \dfrac{h}{x}}\quad[\because \bf{tan\:60^{\circ} = \sqrt{3}}] \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{x = \dfrac{h}{\sqrt{3}}}} \quad \quad Eq..(i) \ \\ \\ \\\end{gathered}$

Hence the distance between base of A and B (in terms of h) is √3/h

Now , by using the tan θ and substituting the values in it, we get :

$\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{tan\:30^{\circ} = \dfrac{h}{100 - x}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - x}}\quad[\because \bf{tan\:30^{\circ} = \dfrac{1}{\sqrt{3}}}] \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100 - x = h\sqrt{3}} \\ \\ \\\end{gathered}$

Now , by substituting the value of x from equation (i) , we get :

$\begin{gathered}:\implies \bf{100 - \dfrac{h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}$

$@\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3} - h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} - h = h\sqrt{3} \times \sqrt{3}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} - h = 3h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} = h + 3h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100\sqrt{3} = 4h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3}}{4} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{25\sqrt{3} = h} \\ \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{h = 25\sqrt{3}}} \\ \\ \\\end{gathered}$

Hence the Height of two towers is 25√3 m.

$\LARGE\bold{Distance\: from\: the \:points\: :}$

$\LARGE\bold{Distance \:between \:A\: and \:B\: :}$

Since, we have taken the base distance as x and we know the value of x in terms of h i.e,

$\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}$

Now, putting the value of h in the above equation , we get :

$\begin{gathered}:\implies \bf{x = \dfrac{25\sqrt{3}}{\sqrt{3}}} \\ \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{x = 25 m} \\ \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{x = 25\:m}} \\ \\ \\\end{gathered}$

Hence, the base distance from A to B is 25.

Distance between B and C :

We know that the distance between B and C is (100 - x) m.

So by putting the value of x in it , we get :

$\begin{gathered}:\implies \bf{100 - x} \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{100 - 25} \\ \\\end{gathered}$

$\begin{gathered}:\implies \bf{75} \\ \\\end{gathered}$

$\begin{gathered}\boxed{\therefore \bf{75\:m}} \\ \\ \\\end{gathered}$

Hence the distance between B and C is 75 m.