Question

From the top of a tower 60m. high, the angles of depression of the top and bottom of a building whose base is in the same straight line with the base of the tower are observed to be 30° and 60° respectively. Find the height of the building.

Answer 1

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,

Height of the tower = $60\: m$

Angle of depression from top of tower to the top of the building = $30^{\circ}$

Angle of depression from the top of the tower to the top of the building = $30^{\circ}$

Height of the building = ? m

$\underline{\large{\textsf{Step-1:}}}$
Picturize the Given data as shown in figure and Express the given data as per figure.

Let A and B be the top and bottom of tower respectively
$\implies \textsf{AB represents the height of the tower}$

AB = 60 cm (Given) ————[1]

and C and D be the top and bottom of building respectively
$\implies \textsf{CD represents the height of the Building}$

Angle of depression from top of tower to the top of the building is $\angle{XAC}$

$\angle{XAC} = 30^{\circ}$(Given)

Angle of depression from the top of the tower to the top of the building is $\angle{XAD}$

$\angle{XAD} = 60^{\circ}$

$\underline{\large{\textsf{Step-2:}}}$
Draw a line Parallel to AX and BX and note alternate angles

Let EC be the line drawn parallel to AX and BD

Here,
When AX and EC are considered,
AC will act as transversal and AX and EC are Parallel to each other.

$\angle{XAC} =\angle{ACE}$ ($\textbf{Alternate\: Angles}$)

Also,
When AX and BD are considered,
AD will act as transversal and AX and BD are Parallel to each other.

$\angle{XAD} =\angle{ADB}$ ($\textbf{Alternate\: Angles}$)

$\underline{\large{\textsf{Step-2:}}}$
Note the Information from the figure

$EC = BD$ ————[2]

$EB = CD$ ————[3]

And
$AB = AE + EB$

Substituting [1]

$AE + EB = 60$

Substituting [3]

$AE + CD = 60$

$\implies \textsf{CD = 60 - AE}$ ————[4]

(CD gives the height of the building, so only $\underline{\textit{AE\: has\: to\: be\: found}}$)

$\underline{\large{\textsf{Step-3:}}}$
Find tan of Angle in $\triangle{ACE}$ (At Point, top of the Bottom of the building) and Simplify.

We have,
$\bigstar \mathbf{tan = \dfrac{Opposite}{Adjacent}}$

$tan\: C = \dfrac{AE}{EC}$

(In $\triangle{ACE}$,
Opposite - AE ; Adjacent - EC)

$tan\: 30^{\circ} = \dfrac{AE}{EC}$

We have,
$tan\: 30^{\circ} = \dfrac{1}{\sqrt{3}}$

$\implies \dfrac{1}{\sqrt{3}}= \dfrac{AE}{EC}$

Substitute [2]

$\implies \dfrac{1}{\sqrt{3}}= \dfrac{AE}{BD}$ ————[4]

$\underline{\large{\textsf{Step-4:}}}$
Find tan of Angle in $\triangle{ADB}$ (At Point, top of the Bottom of the building) and Simplify.

$tan\: D = \dfrac{AB}{BD}$

(In $\triangle{ADB}$,
Opposite - AB ; Adjacent - BD)

$tan\: 60^{\circ} = \dfrac{AB}{BD}$

We have,
$tan\: 60^{\circ} = \sqrt{3}$

$\implies \sqrt{3}= \dfrac{AB}{BD}$

Substitute [1]

$\implies \sqrt{3}= \dfrac{60}{BD}$ ————[5]

$\underline{\large{\textsf{Step-5:}}}$
Express [4] & [5] in terms of BD

From (4)
$\implies BD = \sqrt{3}(AE)$ ————[6]

From (5)
$\implies BD = \dfrac{60}{\sqrt{3}}$ ————[6]

$\underline{\large{\textsf{Step-6:}}}$
Equate [5] & [6]

$\implies \sqrt{3}(AE) = \dfrac{60}{\sqrt{3}}$

$\implies AE = \dfrac{60}{\sqrt{3} \times \sqrt{3}}$

$\implies AE = \dfrac{60}{3}$

$\implies AE = 20$ ————[7]

$\underline{\large{\textsf{Step-7:}}}$
Substitute [7] in [3]

$\implies CD = 60 - 20$

$\therefore$
$\implies CD = 40$

$\bigstar \textsf{The Height of the building is \underline{\textbf{40\: mts}}}$