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Q9:
The angle of elevation of the top of a building from the foot of the tower is $30°$ and the angle of elevation of the top of the tower from the foot of the building is $60°$. If the tower is $50$ m high, find the height of the building.

Solution :

Given:

  • Height of the tower ($CD$) = $50$ m.
  • Angle of elevation of the top of the building ($AB$) from the foot of the tower ($D$) = $30^\circ$.
  • Angle of elevation of the top of the tower ($CD$) from the foot of the building ($B$) = $60^\circ$.

To find:

The height of the building ($AB = h$).

A B C D 30° 60°

Step 1: Define variables and identify triangles.

Let the height of the building $AB = h$ meters.

Let the distance between the foot of the building ($B$) and the foot of the tower ($D$) be $BD = x$ meters.

We have two right-angled triangles: $\triangle ABD$ and $\triangle CDB$.

Step 2: Analyze $\triangle CDB$ to find the distance $x$.

In $\triangle CDB$, the angle of elevation is $\angle CBD = 60^\circ$.

Using the trigonometric ratio tangent: $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan(60^\circ) = \frac{CD}{BD}$

Since $\tan(60^\circ) = \sqrt{3}$ [Trigonometric table value]:

$\sqrt{3} = \frac{50}{x}$

$x = \frac{50}{\sqrt{3}}$ --- (Equation 1)

Step 3: Analyze $\triangle ABD$ to find the height $h$.

In $\triangle ABD$, the angle of elevation is $\angle ADB = 30^\circ$.

Using the trigonometric ratio tangent:

$\tan(30^\circ) = \frac{AB}{BD}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ [Trigonometric table value]:

$\frac{1}{\sqrt{3}} = \frac{h}{x}$

$h = \frac{x}{\sqrt{3}}$ --- (Equation 2)

Step 4: Substitute Equation 1 into Equation 2.

$h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}}$

$h = \frac{50}{\sqrt{3} \times \sqrt{3}}$

$h = \frac{50}{3}$

$h = 16.666...$ meters

Final Answer: The height of the building is $\frac{50}{3}$ m or approximately $16.67$ m.


More Questions from Class 10 Mathematics Applications of Trigonometry EXERCISE 9.1


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