Question

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building​

Answer 1

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$\large{\red{\underline{\tt{Diagram\::-}}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){4.4}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf50 m}\put(7.3,2){\sf50 m}\put(11.1,4){\sf x m}\qbezier(8,1)(10.7,5)(11,5.4)\qbezier(8,3)(10.5,5)(11,5.4)\qbezier(8.65,1)(8.7,1.4)(8.4,1.6)\put(8.7,1.2){\sf60^{\circ} }\qbezier(8.5,3.4)(8.8,3.2)(8.7,3)\put(8.8,3.2){\sf45^{\circ} }\put(7.7,3){\large{A}}\put(7.7,1){\large{B}}\put(11.1,1){\large{C}}\put(11.1,3){\large{D}}\put(11.1,5.3){\large{E}}\put(9.5,0.8){\sf y m}\end{picture}$

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$\large{\red{\underline{\tt{Given\::-}}}}$

The angle of depression of the top and bottom of a 50m high building from the top of the tower are 45° and 60° respectively.

$\large{\red{\underline{\tt{To\:find\::-}}}}$

The height of the tower and the horizontal distance between the tower and the building

$\large{\red{\underline{\tt{Solution\::-}}}}$

• AB = 50m
• EC = (x + 50)m
• XP = y m.

In ∆ EDA,

$\dashrightarrow$$\:$ $\sf tan \:45^{\circ} = \dfrac{ED}{DA}$

$\:\:$

$\dashrightarrow$$\:$$\sf tan\: 45^{\circ} = \dfrac{ED}{BC}$ $\:\:$ ($\because$ DA = BC = y )

$\:\:$

$\dashrightarrow$$\:$$\sf 1 = \dfrac{x}{y}$

$\:\:$

$\dashrightarrow$$\:$$\sf y = x$ -------(1)

$\:\:$

In ∆ECB,

$\:\:$

$\dashrightarrow$$\:$$\sf tan\:60^{\circ} = \dfrac{EC}{BC}$

$\:\:$

$\dashrightarrow$$\:$$\sf \sqrt{3} = \dfrac{x + 50}{y}$

$\:\:$

$\dashrightarrow$$\:$$\sf \sqrt{3} = \dfrac{x + 50}{x}$ $\:\:$ ($\because$ from equation (1))

$\:\:$

$\dashrightarrow$$\:$$\sf \sqrt{3}x = x + 50$

$\:\:$

$\dashrightarrow$$\:$$\sf \sqrt{3}x - x = 50$

$\:\:$

$\dashrightarrow$$\:$$\sf x \left(\sqrt{3} - 1 \right) = 50$

$\:\:$

$\dashrightarrow$$\:$$\sf x = \dfrac{50}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$\:\:$

$\dashrightarrow$$\:$$\sf x = \dfrac{50\left(\sqrt{3} + 1\right)}{3 - 1}$

$\:\:$

$\dashrightarrow$$\:$$\sf x = \dfrac{50\left(1.73 + 1\right)}{2}$

$\:\:$

$\dashrightarrow$$\:$$\sf x = 68.25 = y$

$\:\:$

$\dagger$$\:\:$Height of the tower :-

$\:\:$

$\dashrightarrow$$\:$$\sf Tower = EC$

$\:\:$

$\dashrightarrow$$\:$$\sf Tower = x + 50$

$\:\:$

$\dashrightarrow$$\:$$\sf Tower = 68.25 + 50$

$\:\:$

$\dashrightarrow$$\:\:$$\underline{\boxed{\gray{\textbf{Tower = 118.25\:m}}}}$ $\orange\bigstar$

$\therefore$ The height of the tower is 118.25 m. and the horizontal distance between the tower and building is 68.25 m.

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Answer 2

★ Figure refer to attachment

Given

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively.

Find out

Find the height of the tower and the horizontal distance between the tower and the building.

Solution

• Building = ED = BC = 50m
• Tower = AC
• BE = CD
• ∠FAE = 45° = ∠AEB
• ∠FAD = 60° = ∠ADC
• AB = (AC - 50)

In ∆ABE

$\implies\sf tan45°=\dfrac{AB}{BE} \\ \\ \\ \implies\sf 1=\dfrac{AC-50}{BE} \\ \\ \\ \implies\sf BE=(AC-50) \\ \\ \\ \bigstar\bf{\red{BE=CD}}\;\;\;(Given) \\ \\ \\ \implies\sf CD=(AC-50) ----(i)$

In ∆ACD

$\implies\sf tan60°=\dfrac{AC}{CD} \\ \\ \\ \implies\sf \sqrt{3}=\dfrac{AC}{CD} \\ \\ \\ \implies\sf AC=\sqrt{3}CD---(ii)$

From (i)

➞ CD = (AC - 50)

➞ CD = √3CD - 50 [Using (ii)]

➞ 50 = √3CD - CD

➞ 50 = CD(√3 - 1)

★ Rationalize its denominator

➞ CD = 50/(√3 - 1) × (√3 + 1)/(√3 + 1)

➞ CD = 50(√3 + 1)/(√3)² - (1)²

➞ CD = 50(√3 + 1)/3 - 1

➞ CD = 50(√3 + 1)/2

➞ CD = 25(1.73 + 1)

➞ CD = 25 × 2.73

➞CD = 68.25 m

Now, put the value of CD in eqⁿ(ii)

➞ AC = √3CD

➞ AC = 1.73 × 68.25

➞ AC = 118.07m

Hence,

• Height of tower = AC = 118.07 m
• Distance between building and tower = CD = 68.25

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