Question

A multi-storey building of 40 m height is standing in front of a tower at a horizontal distance of 20√3m. The angle of elevation of the top of the tower observed from the foot of the building is of 60° Find the angle of elevation of the top of the tower from the top of the building.​

Answer 1

Given:

Height of building = 40 m

Distance between the towers = 20√3 m

Angle of elevation to the top of the tower from bottom of the buliding = 60°

To Find:

Angle of elevation of the top of the tower from top of buliding

Solution:

Let the height of the tower be AC

Let the height of buliding be ED = 40 m

Let distance between the towers be CD = 20√3 m

First we have to find the height of the tower

Consider Δ ABD

➝ tan 60 = AC/CD

➝ √3 = AC/20√3

➝ AC = 20 √3 × √3

➝ AC = 60 m

Now consider Δ ABE

➝ tan θ = AB/BE

We know that

➝ AB = 60 - 40 = 20 m

➝ BE = CD = 20√3 m

Hence

➝ tan θ = 20/20√3

➝ tan θ = 1/√3

➝ tan θ = tan 30°

Hence

➝ θ = 30°

Therefore angle of elevation to the top of tower from top of building is 30°

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Answer 2

Given :-

$~$

• Height of building = $40$m
• Distance between the towers = $20√3$m
• Angle of elevation to the top of the tower from bottom of the building = $60°$

$~$

To find :-

$~$

• Angle of elevation of the top of the tower from top of building.

$~$

Solution :-

$~$

• Let the height of the tower be AC

• Let the height of the building be ED = $40$m

• Let distance between the tower be CD = $20√3$m

$~$

★ First we have to find the height of the tower :

$~$

$~~~~~~~~~~~~~~~~~~~~$ Consider ∆ ABD

$~$

:$\implies$ tan 60 = $\large{\sf{ \frac{AC}{CD} }}$

$~$

$~~~~~$:$\implies$ $√3$ = $\large{\sf{\frac{AC}{20√3}}}$

$~$

$~~~~~~~~~~$:$\implies$ AC = $20~√3~×~√3$

$~$

$~~~~~~~~~~~~~~~$:$\implies$ $\large{\underline{\boxed{\bf{\pink{AC~=~60~m}}}}}$★

$~$

$~~~~~~~~~~~~~$ Now, Consider ∆ ABE

$~$

• $\implies$ tan ø = $\large{\sf{\frac{AB}{BE}}}$

$~$

★ As we know that,

$~$

• $\implies$ AB = $60 - 40 = 20$m

$~$

• $\implies$ BE = CD = $20√3$m

$~$

$\large\dag$ Hence,

$~$

$\implies$ tan ø = $\large{\sf{\frac{20}{20√3}}}$

$~$

$~~~~~$:$\implies$ tan ø = $\large{\sf{\frac{1}{√3}}}$

$~$

$~~~~~~~~~~$:$\implies$ tan ø = $\large{\underline{\boxed{\bf{\green{tan~30°}}}}}$★

$~$

$\large\dag$ Hence Proved

$~$

• Therefore angle of elevation to the top of tower from top of building is ${\bf{30°}}$ $\large{\bf\green{✓}}$