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Q11:

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60°$. From another point $20$ m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is $30°$ (see Fig. 9.12). Find the height of the tower and the width of the canal.

Solution :

Given:

A TV tower $AB$ stands vertically on the bank of a canal. Let $BC$ be the width of the canal. From a point $C$ on the other bank, the angle of elevation of the top of the tower $A$ is $60^\circ$. From another point $D$, which is $20\text{ m}$ away from $C$ on the line joining $C$ to the foot of the tower $B$, the angle of elevation of the top of the tower $A$ is $30^\circ$.

To Find:

1. The height of the tower ($h = AB$).

2. The width of the canal ($x = BC$).

A B C D 60° 30° 20 m

Step 1: Defining Variables and Assumptions

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

Let the width of the canal $BC = x$ meters.

The distance $BD = BC + CD = x + 20$ meters.

Step 2: Analyzing Triangle ABC

In the right-angled triangle $\triangle ABC$, the angle of elevation at $C$ is $60^\circ$.

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

$\tan(60^\circ) = \frac{AB}{BC}$

Since $\tan(60^\circ) = \sqrt{3}$:

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

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

Step 3: Analyzing Triangle ABD

In the right-angled triangle $\triangle ABD$, the angle of elevation at $D$ is $30^\circ$.

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

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

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ and $BD = x + 20$:

$\frac{1}{\sqrt{3}} = \frac{h}{x + 20}$

$x + 20 = h\sqrt{3}$ --- (Equation 2)

Step 4: Solving the System of Equations

Substitute the value of $h$ from Equation 1 into Equation 2:

$x + 20 = (x\sqrt{3})\sqrt{3}$

$x + 20 = 3x$ [Since $\sqrt{3} \times \sqrt{3} = 3$]

$20 = 3x - x$

$20 = 2x$

$x = 10$

Thus, the width of the canal is $10\text{ m}$.

Step 5: Calculating the Height of the Tower

Substitute $x = 10$ back into Equation 1:

$h = 10\sqrt{3}$

Using $\sqrt{3} \approx 1.732$:

$h = 10 \times 1.732 = 17.32\text{ m}$

Final Answer: The height of the tower is $10\sqrt{3}\text{ m}$ (or approximately $17.32\text{ m}$) and the width of the canal is $10\text{ m}$.


More Questions from Class 10 Mathematics Applications of Trigonometry EXERCISE 9.1


CBSE Solutions for Class 10 Mathematics Applications of Trigonometry


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