Question

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide.From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles​

Answer 1

$\huge{\orange{\underline{\red{\mathbb{Question}}}}}$

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Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide.From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

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✪ $\huge{\orange{\underline{\red{\mathbb{Solution}}}}}$✪

$From\: the\: figure$

$\: \: \: \tan60° = \frac{h}{x} \: \: \sqrt{3 } = \frac{h}{x} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = \sqrt{3} x \: \: ........ \: (1) \\ \\ also \: \tan30° = \frac{h}{120 - x} \: \: \frac{1}{ \sqrt{3} } = \frac{h}{120 - x} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = \frac{120 \times x}{ \sqrt{3} } \: \: ....... \: (2) \\ \\ from \: (1) \: and \: (2) \\ \\ ⇒\sqrt{3} x = \frac{120 - x}{ \sqrt{3} } \\ ⇒ \sqrt{3} . \sqrt{3} x = 120 - x \\ ⇒3x = 120 - x \\ ⇒3x + x = 120 \\ ⇒4x = 120 \\ \\ ⇒x = \frac{120}{4} = 30ft. \\ \\ now \: h = \sqrt{3} x = \sqrt{3} \times 30 \\ = 1.732 \times 30 \\ = 51.960 \: feet$

$∴\: distance \:of\:the\:poles=30\:ft.\:and$

$120-30\:fts\:=$$\red{90\:ft.}$

$height\:of\:each\:pole=$$\red{51.96\:ft.}$

Answer 2

Answer:

Height of the pole= 51.96m

Distance of given point from poles= 30m and 90m

☸$\tt\bold{\underline{Solution}}$☸

Let AB and CD be two towers of equal height 'h'. DB be the road of width 120m. E be the point on the road such that DE=x, then EB=120-x.

$\sf\bold{\underline{Diagram:}}$ Please see the provided attachment.

Things to be known before solving question:

$\sf\rightarrow{\tan{\theta}=\dfrac{Perpendicular}{Base}}$

$\sf\rightarrow{\tan{30}^{\circ}=\dfrac{1}{\sqrt{3}}}$

$\rightarrow\sf{\tan{60}^{\circ}={\sqrt{3}}}$

$\rightarrow\sf{\sqrt{3}=1.732}$

In ΔABE,

$\rightarrow\sf{\tan{30}^{\circ}=\dfrac{AB}{BE}}$

$\rightarrow\sf{\dfrac{1}{\sqrt{3}}=\dfrac{h}{120-x}}$

$\rightarrow\sf{h=\dfrac{120-x}{\sqrt{3}}\ \ \ \ \ \ \ \ \ \longrightarrow\ (1)}$

In ΔCDE,

$\rightarrow\sf{\tan{60}^{\circ}=\dfrac{CD}{DE}}$

$\rightarrow\sf{{\sqrt{3}}=\dfrac{h}{x}}$

$\rightarrow\sf{h={\sqrt{3}x}\ \ \ \ \ \ \ \ \ \ \ \ \ \longrightarrow\ (2)}$

From (1) and (2)

$\rightarrow\sf{{\dfrac{120-x}{\sqrt{3}}={\sqrt{3}x}}}$

$\rightarrow\sf{120=x+3x}$

$\rightarrow\sf{120=4x}$

$\rightarrow\sf{x=\dfrac{120}{4}}$

$\rightarrow\sf{\purple{x=30}}$

☛ So, Distance of point E from:

• $\tt\green{ {Pole\:AB=120-x=120-30=90m}}$
• $\tt\green{ {Pole\:CD=x=30m}}$

☛ Also, height of both the poles

• $\tt{\green{ h=\sqrt{3}x}}$
• $\tt{\green{ h=30\sqrt{3}}}$
• $\tt{\green{h=30\times1.732}}$
• $\tt{\green{h=51.96m}}$