Question

the angle of elevation of the top of a tower from a point on the ground which is 3cm away from the foot of the tower is 30° . find the height of the tower ​

Answer 1

In the above Question , the following information is given -

The angle of elevation of the top of a tower from a point on the ground which is 3cm away from the foot of the tower is 30 °

To find -

Find the height of the tower .

Solution -

Here , we have to find the height of the tower .

So , let us assume that the required height of the tower is x metres.

Now ,

We know that :

Tan theta = [ Perpendicular / Base ]

Now, refer to the attached figure .

Here, the tower is denoted by OA .

The distance from the foot of the tower , OP is 3 cm .

The required Angle if elevation , angle OAP is 30°

Now ,

Tan OAP -

=> [ OP / OA ]

Substituting the given value -

Tan OAP

=> [ x / 3 ]

Now ,

Angle OAP = 30°

Tan 30°

=> { 1 / √ 3 }

=> { √ 3 / 3 }

Thus ,

[ x / 3 ] = [ √ 3 / 3 ]

Thus , x = √ 3 cm .

But, x is the height of the tower .

So ,

The required height of the tower is √ 3 cm .

____________

Additional Information -

Sin theta = [ Perpendicular / Hypotenuse ]

Cos theta = [ Base / Hypotenuse ]

________

Answer 2

$\huge\sf\pink{Answer}$

☞ Height of the tower is 10√3 m

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ The angle of elevation at the top of a river from a point on ground is 3 cm.

✭ Angle of elevation from the top of the tower = 30°

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

➳ The height of tower?

$\rule{110}1$

$\huge\sf\purple{Steps}$

➢ Let the height of tower be AB

➢ And BC = 30 cm

➢ Angle of elevation from the top of a tower to a point on the ground is 30°

In ∆ABC,

$\twoheadrightarrow\sf{\dfrac{P}{B}\:=\:tan30^{\circ}}$

$\twoheadrightarrow\sf{\dfrac{AB}{BC}\:=\:tan30^{\circ}}$

We know that,

BC = 30 m

tan30° = 1/√3

Substituting the given values,

$\dashrightarrow\sf{\dfrac{AB}{30}\:=\:\dfrac{1}{\sqrt{3}}}$

$\dashrightarrow\sf{AB\sqrt{3}\:=1(30)}$

$\dashrightarrow\sf{AB\sqrt{3}\:=\:30}$

Finding AB

Take √3 on right hand side.

$\dashrightarrow\sf{AB\:=\:\dfrac{30}{\sqrt{3}}}$

$\dashrightarrow\sf{AB\:=\:\dfrac{30}{\sqrt{3}}\:\times\:\dfrac{\sqrt{3}}{\sqrt{3}}}$

$\sf\dashrightarrow \sqrt{3}× \sqrt{3 }= (\sqrt{3})^2 = 3$

$\dashrightarrow\sf{AB\:=\:\dfrac{30\sqrt{3}}{3}}$

$\orange{\dashrightarrow\sf{AB\:=\:10\sqrt{3} \ m}}$

$\rule{170}3$