Question

The angle of elevation of the top of a building from the foot of the tower is 30 degree and the angle of elevation of the top of the tower from the foot of the building is 60. If the building height is 20m then find the height of the tower

Answer 1

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Answer 2

Given :

• The angle of elevation of the top of a building from foot of tower is 30°
• The angle of elevation of the top of the tower from foot of building is 60°

To Find :

• Height of the tower.

Solution :

Consider, ΔABC.

AB = 20 m = Height of building

BC = Distance between tower and building.

$\bold{\theta\:=\:60\:^\circ}$

Using the trigonometric function, tan.

We know,

$\sf{tan\:\theta\:=\:\dfrac{Opposite\:side\:to\:\angle\:\theta}{Adjacent\:side\:to\:\angle\:\theta}}$

Block in the data,

$\sf{\longrightarrow{tan\:30\:^\circ\:=\:\dfrac{AB}{BC}}}$

$\sf{\longrightarrow{\dfrac{1}{\sqrt{3}}\:=\:\dfrac{20}{BC}}}$

$\bold{\big[\because\:tan\:^\circ\:=\:\dfrac{1}{\sqrt{3}}\big]}$

$\sf{\longrightarrow{BC\:=\:20\:\sqrt{3}\:\:\:\:(1)}}$

Now, consider ΔBCD.

DC = x m

BC = Distance between tower and building.

$\bold{\theta\:=\:60\:^\circ}$

Using the same trigonometric function, tan.

$\sf{\longrightarrow{tan\:\theta\:=\:\dfrac{x}{BC}}}$

$\sf{\longrightarrow{tan\:60\:^\circ\:=\:\dfrac{x}{BC}}}$

$\sf{\longrightarrow{\sqrt{3}=\dfrac{x}{BC}}}$

$\bold{\big[\because\:tan\:60\:^\circ\:=\:\sqrt{3}\big]}$

$\sf{\longrightarrow{BC\:\sqrt{3}=\:x}}$

$\sf{\longrightarrow{BC\:=\:\dfrac{x}{\sqrt{3}}\:\:\:\:(2)}}$

Comparing equation (1) and (2),

$\sf{20\:\sqrt{3}\:=\:\dfrac{x}{\sqrt{3}}}$

$\sf{\longrightarrow{20\:\sqrt{3}\:\times\:\sqrt{3}=x}}$

$\sf{\longrightarrow{20\:\times\:3=x}}$

$\sf{\longrightarrow{60=x}}$

$\large{\boxed{\bold{\red{Height\:of\:tower\:=DC=\:60\:m}}}}$