Question

The angle of elevation and the angle of depression of the top and foot of the monument, when observed from a point on the roof of a five-storied building of Mihir are 60° and 30° respectively. If the height of building is 16 metre, what will be the height of the monument ??????​

Answer 1

Answer:

Refer to the Attachment for the diagram.

Important Interpretations from the diagram:

• AB = CD = 16 m
• AD = BC (Parallel and Equal)
• DE = 'x'
• Height of Monument = CE = (x+CD)
• ∠DAC = ∠ACB = 30° (Adjacent Angles)

From the interpretation, we get:

⇒ Height of monument = x + 16 m.

Consider ΔADE,

$\implies Tan\:60 = \dfrac{ED}{AD}\\\\\\\implies \sqrt{3} = \dfrac{x}{AD}\\\\\\\implies \boxed{ \bf{ x = AD\sqrt{3}}}$

Consider ΔABC,

$\implies Tan\:30 = \dfrac{AB}{BC}\\\\\\\implies \dfrac{1}{\sqrt{3}} = \dfrac{16}{AD}\\\\\\\implies \boxed{ \bf{ AD = 16\sqrt{3}}}$

Substituting the value of AD, we get:

⇒ x = AD√3

⇒ x = ( 16√3 ) × √3

⇒ x = 16 × 3

⇒ x = 48 m.

Hence the height of the monument is:

⇒ (x + 16) = (48 + 16) = 64 m

Answer 2

Given that , The angle of elevation and the angle of depression of the top and foot of the monument, when observed from a point on the roof of a five-storied building of Mihir are 60° and 30° respectively & the height of building is 16 metres .

Exigency To Find : The Height of Monument ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's say that DE be the Height of Building & AC be the Height of Building. [ Refer to attachment ]

Now ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ In Right Angled Triangle AEB :

$\qquad \dashrightarrow \sf Tan \:60^\circ \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: AEB \:)} \:}{ Base \:_{(\:\triangle \: AEB \:)}}$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ \:\:Tan \:60^\circ \:=\:\sqrt{3} }\bigg\rgroup$

$\qquad \dashrightarrow \sf Tan \:60^\circ \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: AEB \:)} \:}{ Base \:_{(\:\triangle \: AEB \:)}}$

$\qquad \dashrightarrow \sf \sqrt{3} \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: AEB \:)} \:}{ Base \:_{(\:\triangle \: AEB \:)}}$

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

$\qquad \dashrightarrow \sf \sqrt{3} \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: AEB \:)} \:}{ Base \:_{(\:\triangle \: AEB \:)}}$

Here ,

• Perpendicular of Triangle AEB = AB = a m
• Base of Triangle AEB = EB

$\qquad \dashrightarrow \sf \sqrt{3} \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: AEB \:)} \:}{ Base \:_{(\:\triangle \: AEB \:)}}$

$\qquad \dashrightarrow \sf \sqrt{3} \:\: = \: \dfrac{ a \:}{ EB }$

$\qquad \dashrightarrow \sf a \:\: = \: EB \sqrt{3}$

$\qquad \dashrightarrow \sf a \:\: = \: EB \sqrt{3}\qquad \:\bigg\lgroup \sf{ \:\:Eq^n \:I }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ In Right Angled Triangle EDC :

$\qquad \dashrightarrow \sf Tan \:30^\circ \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: EDC \:)} \:}{ Base \:_{(\:\triangle \: EDC\:)}}$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ \:\:Tan \:30^\circ \:=\:\dfrac{1}{\sqrt{3}} }\bigg\rgroup$

$\qquad \dashrightarrow \sf Tan \:30^\circ \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: EDC \:)} \:}{ Base \:_{(\:\triangle \: EDC \:)}}$

$\qquad \dashrightarrow \sf \dfrac{1}{\sqrt{3}} \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: EDC\:)} \:}{ Base \:_{(\:\triangle \: EDC\:)}}$

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

$\qquad \dashrightarrow \sf \dfrac{1}{ \sqrt{3}} \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: EDC\:)} \:}{ Base \:_{(\:\triangle \: EDC \:)}}$

Here ,

• Perpendicular of Triangle EDC = ED
• Base of Triangle EDC = DC

$\qquad \dashrightarrow \sf \dfrac{1}{\sqrt{3}} \:\: = \: \dfrac{ Perpendicular\:_{(\:\triangle \: EDC \:)} \:}{ Base \:_{(\:\triangle \: EDC \:)}}$

$\qquad \dashrightarrow \sf \dfrac{1}{\sqrt{3}} \:\: = \: \dfrac{ ED\:}{ DC }$

• ED = 16 m [ Height of Building ]
• DC = EB [ Parallel Sides ]

$\qquad \dashrightarrow \sf \dfrac{1}{\sqrt{3}} \:\: = \: \dfrac{ ED\:}{ DC }$

$\qquad \dashrightarrow \sf \dfrac{1}{\sqrt{3}} \:\: = \: \dfrac{ 16\:}{ EB }$

$\qquad \dashrightarrow \sf EB \:\: = \: 16\sqrt{3}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Value \:of \:EB \:in \:Eq^n \: I \::}}$

$\qquad \dashrightarrow \sf a \:\: = \: EB \sqrt{3}\qquad \:\bigg\lgroup \sf{ \:\:Eq^n \:I }\bigg\rgroup$

$\qquad \qquad \leadsto \sf EB \:\: = \: 16\sqrt{3}$

$\qquad \dashrightarrow \sf a \:\: = \: EB \sqrt{3}\qquad \:$

$\qquad \dashrightarrow \sf a \:\: = \: \Big\{ 16 \sqrt{3} \Big\} \sqrt{3}\qquad \:$

$\qquad \dashrightarrow \sf a \:\: = \: 16\times 3 \qquad \:$

$\qquad \dashrightarrow \sf a \:\: = \: 48 \: \qquad \:$

$\qquad \dashrightarrow\underline {\boxed{\pmb{\frak{\pink{ a \:=\:48 \: m \:}}}}}\:\:\bigstar$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ Height _{(Monument\:)} = \:AB \: + \:BC \:}\bigg\rgroup$

$\qquad \dashrightarrow \sf Height _{(Monument\:)} = \:AB \: + \:BC \:$

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

$\qquad \dashrightarrow \sf Height _{(Monument\:)} = \:AB \: + \:BC \:$

Here ,

• AB = a m = 48 m
• BC = 16 m

$\qquad \dashrightarrow \sf Height _{(Monument\:)} = \:AB \: + \:BC \:$

$\qquad \dashrightarrow \sf Height _{(Monument\:)} = \:48 \: + \:16 \:$

$\qquad \dashrightarrow \sf Height _{(Monument\:)} = \:64 \: m \:$

$\qquad \dashrightarrow\underline {\boxed{\pmb{\frak{\pink{ Height _{(Monument\:)} = \:64 \: m \:}}}}}\:\:\bigstar$

$\qquad \therefore \:\underline {\sf Hence, \:The \:Height \:of \:Monument \:is \:\pmb{\bf 64 \:m \:}\:.}$