Question

A man finds the angle of elevation of a tower to be 60°. When he walks away at a distance of 50m he finds the angle to be 30°, then the height of the tower is​

Answer 1

✬ Height = 25√3 ✬

Step-by-step explanation:

Given:

• Angle of elevation of tower is 60°.
• After walking 50 m away from tower it changes to 30°.

To Find:

• What is height of tower ?

Solution: Let the height of tower AB be h m and distance between the man initial position and foot of tower be x m. Therefore,

• AB (perpendicular) = h

• DB (base) = x

• ∠ABD = 90°

• ∠ADB (angle of elevation) = 60°

[ Now he walked 50 m away from point D. Let he is at now point C. ]

• DC = 50 m

• BC = DB + DC

• ∠ACB (angle of elevation) = 30°

In ∆ABD , using tanθ

★ tanθ = Perpendicular/Base ★

➯ tan60° = AB/DB

➯ √3 = h/x

➯ √3x = hㅤㅤㅤㅤㅤ(eqⁿ i)

Now in ∆ABC , again by tanθ

➯ tan30° = AB/BC

➯ 1/√3 = h/x + 50

➯ x + 50 = √3h

➯ x + 50 = √3 × √3x

➯ 50 = 3x – x

➯ 50/2 = x

➯ 25 m = x

Putting the value of x in eqⁿ (i)

$\implies{\rm }$ √3x = h

$\implies{\rm }$ √3 × 25 = h

$\implies{\rm }$ 25√3 = h

Hence, the height of tower is 25√3 m.

Answer 2

Given : The angle of elevation of a tower to be 60° & He walks away at a distance of 50m he finds the angle to be 30° .

Exigency To Find : The Height of the tower .

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[ Note : Kindly refer to given attachment ( image ) . ]

❍ Let's Consider the Height of Tower AB be h and his initial ( or first ) position DB be x .

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ Tan\:\theta \:: \dfrac{Perpendicular}{Base} }\bigg\rgroup$

Now ,

⠀⠀⠀⠀⠀In $\triangle$ ABD .

$\qquad:\implies \sf Tan 30^\circ \:\:= \: \dfrac { AB }{BC }$

⠀⠀⠀⠀⠀OR,

$\qquad:\implies \sf Tan 30^\circ \:\:= \: \dfrac { h}{x + 50 }$

$\qquad:\implies \bf Tan 30^\circ \:\:= \: \dfrac { h}{x + 50 } \qquad \longrightarrow \:\:Eq.1\:$

Now ,

⠀⠀⠀⠀⠀In $\triangle$ ABC .

$\qquad:\implies \sf Tan 60^\circ \:\:= \: \dfrac { AB }{DB }$

⠀⠀⠀⠀⠀OR,

$\qquad:\implies \sf Tan \:60^\circ \:\:= \: \dfrac { h}{x }$

$\qquad:\implies \bf Tan \:60^\circ \:\:= \: \dfrac { h}{x } \qquad \longrightarrow \:\:Eq.2\:$

⠀⠀⠀⠀⠀Now from Eq.1 & Eq.2 :

$\qquad:\implies \bf Tan 30^\circ \:\:= \: \dfrac { h}{x + 50 } \qquad \longrightarrow \:\:Eq.1\:$

$\qquad:\implies \bf Tan \:60^\circ \:\:= \: \dfrac { h}{x } \qquad \longrightarrow \:\:Eq.2\:$

$\qquad:\implies \sf \dfrac{Tan 60^\circ}{Tan 30^\circ} \:\:= \: \dfrac{\dfrac{h}{x}}{\dfrac { h}{x + 50 }} \:$

$\qquad:\implies \sf \dfrac{Tan 60^\circ}{Tan 30^\circ} \:\:= \: \dfrac{\dfrac{\cancel{h}}{x}}{\dfrac { \cancel {h}}{x + 50 }} \:$

$\qquad:\implies \sf \dfrac{Tan 60^\circ}{Tan 30^\circ} \:\:= \: \dfrac{x}{x + 50 } \:$

Now ,

⠀⠀⠀⠀⠀━━ $Tan \:60^\circ \:= \:\sqrt {3} \:\& \:\: Tan \:30^\circ \:= \:\dfrac{1}{\sqrt{3}}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf \dfrac{Tan 60^\circ}{Tan 30^\circ} \:\:= \: \dfrac{x}{x + 50 } \:$

$\qquad:\implies \sf \dfrac{\sqrt{3}}{\dfrac{1}{\sqrt{3}}} \:\:= \: \dfrac{x}{x + 50 } \:$

$\qquad:\implies \sf x + 50 =\:3x \: \:$

$\qquad:\implies \sf 50 = \:3x - x \: \:$

$\qquad:\implies \sf 2x = 50 \: \:$

$\qquad:\implies \sf x = \cancel {\dfrac{50}{2}}\: \:$

$\qquad \longmapsto \frak{\underline{\purple{\:x = 25 \:m \: }} }\bigstar$

Now ,

⠀⠀⠀⠀⠀━━ From Eq.2 :

$\qquad:\implies \bf Tan \:60^\circ \:\:= \: \dfrac { h}{x } \qquad \longrightarrow \:\:Eq.2\:$

$\qquad:\implies \sf Tan \:60^\circ \:\:= \: \dfrac { h}{x } \qquad \:$

$\qquad:\implies \sf x \times Tan \:60^\circ \:\:= \: h \qquad \:$

$\qquad:\implies \sf 25 \times Tan \:60^\circ \:\:= \: h \qquad \:$

$\qquad:\implies \sf 25 \times \sqrt {3}\:\:= \: h \qquad \:$

$\qquad \longmapsto \frak{\underline{\purple{\:h = 25\sqrt {3}\:m\:\:}} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Height \:of\:Tower \:is\:\bf{25\sqrt{3} \:m \:}}}}$

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