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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.
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 ($x = BC$).
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
- Q1: A circus artist is climbing a $20$ m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30°$ (see Fig. 9.11).
- Q2: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30°$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8$ m. Find the height of the tree.
- Q3: A contractor plans to install two slides for the children to play in a park. For the children below the age of $5$ years, she prefers to have a slide whose top is at a height of $1.5$ m, and is inclined at an angle of $30°$ to the ground, whereas for elder children, she wants to have a steep slide at a height of $3$m, and inclined at an angle of $60°$ to the ground. What should be the length of the slide in each case?
- Q4: The angle of elevation of the top of a tower from a point on the ground, which is $30$ m away from the foot of the tower, is $30°$. Find the height of the tower.
- Q5: A kite is flying at a height of $60$ m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60°$. Find the length of the string, assuming that there is no slack in the string.
- Q6: A $1.5$ m tall boy is standing at some distance from a $30$ m tall building. The angle of elevation from his eyes to the top of the building increases from $30°$ to $60°$ as he walks towards the building. Find the distance he walked towards the building.
- Q7: From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20$ m high building are $45°$ and $60°$ respectively. Find the height of the tower.
- Q8: A statue, $1.6$ m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60°$ and from the same point the angle of elevation of the top of the pedestal is $45°$. Find the height of the pedestal.
- 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.
- Q10: Two poles of equal heights are standing opposite each other on either side of the road, which is $80$ m wide. From a point between them on the road, the angles of elevation of the top of the poles are $60°$ and $30°$, respectively. Find the height of the poles and the distances of the point from the poles.
- Q12: From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $60°$ and the angle of depression of its foot is $45°$. Determine the height of the tower.
- Q13: As observed from the top of a $75$ m high lighthouse from the sea-level, the angles of depression of two ships are $30°$ and $45°$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
- Q14: A $1.2$ m tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2$ m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60°$. After some time, the angle of elevation reduces to $30°$ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.
- Q15: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30°$, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be $60°$. Find the time taken by the car to reach the foot of the tower from this point.
CBSE Solutions for Class 10 Mathematics Applications of Trigonometry
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