9.1-Heights and Distances
9.1-Heights and Distances Important Formulae
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Grade 10 → Math → Some Applications of Trigonometry → 9.1-Heights and Distances
- Identify line of sight in order to determine angle of elevation and angle of depression.
- Apply trigonometric ratios (of specific angles) in order to determine heights and distances of the objects.
Heights and distances problems involve the application of trigonometric concepts to calculate the heights of objects and distances between points that are not directly measurable. These problems are typically solved using right-angled triangles and trigonometric ratios.
1. Basic Concepts
When dealing with heights and distances, we often visualize scenarios where an observer looks at an object at a certain height from a specific distance. The angle of elevation and angle of depression are key concepts in these problems:
- Angle of Elevation: The angle formed by the line of sight when looking up at an object. For instance, if you look at the top of a tree, the angle of elevation is formed above the horizontal line.
- Angle of Depression: The angle formed by the line of sight when looking down at an object. For example, if you are standing on a hill and looking down at a car, the angle of depression is formed below the horizontal line.
2. Trigonometric Ratios
To solve heights and distances problems, we use the following trigonometric ratios:
- Sine: $ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} $
- Cosine: $ \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} $
- Tangent: $ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $
3. Solving Height Problems
To find the height of an object, we can use the angle of elevation. For example, if a person is standing 30 meters away from a tree and the angle of elevation to the top of the tree is $60^\circ$, we can find the height $h$ of the tree as follows:
Using the tangent ratio:
$$ \tan(60^\circ) = \frac{h}{30} $$
Rearranging gives:
$$ h = 30 \times \tan(60^\circ) $$
Calculating this yields:
$$ h = 30 \times \sqrt{3} \approx 51.96 \, \text{meters} $$
4. Solving Distance Problems
To find the distance to an object when its height is known, we can use the angle of depression. For instance, if a person standing on a tower observes a car at an angle of depression of $45^\circ$ and the height of the tower is 40 meters, we can find the distance $d$ from the base of the tower to the car:
Using the tangent ratio:
$$ \tan(45^\circ) = \frac{40}{d} $$
Since $\tan(45^\circ) = 1$, we have:
$$ 1 = \frac{40}{d} $$
Thus:
$$ d = 40 \, \text{meters} $$
5. Applications in Real Life
Heights and distances problems are applicable in various real-life scenarios, including:
- Architecture: Calculating the height of buildings and structures.
- Aviation: Determining the altitude of aircraft.
- Navigation: Finding distances between locations using angles from landmarks.
6. Important Considerations
When solving heights and distances problems, keep the following points in mind:
- Always ensure that the angles are measured from a horizontal line.
- Draw diagrams to visualize the scenario clearly.
- Use the correct trigonometric ratios based on the known sides and angles.
7. Practice Problems
To reinforce understanding, it's essential to practice various heights and distances problems. Here are some sample problems:
- Find the height of a tower if the angle of elevation from a point 50 meters away is $30^\circ$.
- Determine the distance from a point on the ground to the base of a building if the height of the building is 60 meters and the angle of depression from the top is $45^\circ$.
Solving these types of problems helps strengthen the understanding of trigonometric applications in real-world contexts.
Standard problem situation in trigonometry.
MikeRun, CC BY-SA 4.0, via Wikimedia Commons
Solved Example: 9-1-01
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).
Solution:
Height of the pole = 10 m
Solved Example: 9-1-02
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.
Solution:
Total height of the tree (before storm) = 8$\sqrt{3}$
Solved Example: 9-1-03
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 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
To find the length of each slide, we can use trigonometry, specifically the sine function. The length of the slide can be calculated using the formula:
\[L = \frac{h}{\sin(\theta)}\]
where \( L \) is the length of the slide, \( h \) is the height of the slide, and \( \theta \) is the angle of inclination.
For the slide for children below the age of 5 years:
- Height (\( h \)) = 1.5 m- Angle (\( \theta \)) = 30°
Calculating the length: \[L_1 = \frac{1.5}{\sin(30^\circ)}\] Since \( \sin(30^\circ) = 0.5 \): \[L_1 = \frac{1.5}{0.5} = 3 \text{ m}\]
For the slide for elder children:
- Height (\( h \)) = 3 m- Angle (\( \theta \)) = 60°
Calculating the length: \[L_2 = \frac{3}{\sin(60^\circ)}\] Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \): \[L_2 = \frac{3}{\frac{\sqrt{3}}{2}} = \frac{3 \times 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} \approx 3.46 \text{ m}\]
Summary of the lengths of the slides:
- Length for the slide for children below 5 years: 3 m- Length for the slide for elder children: approximately 3.46 m
Solved Example: 9-1-04
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.
Solution:
To find the height of the tower, we can use the trigonometric relationship involving the tangent of the angle of elevation.
Given:
- Distance from the point on the ground to the foot of the tower (\(d\)) = 30 m
- Angle of elevation (\(\theta\)) = 30°
Let the height of the tower be \(h\).
From the definition of tangent in a right triangle:
\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]
Here, the opposite side is the height of the tower (\(h\)), and the adjacent side is the distance from the foot of the tower to the point (\(d\)).
Thus, we have:
\[\tan(30^\circ) = \frac{h}{30}\]
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). Substituting this value in:
\[\frac{1}{\sqrt{3}} = \frac{h}{30}\]
To solve for \(h\), we can rearrange the equation:
\[h = 30 \times \frac{1}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3}\]
Now, calculating the numerical value:
\[10\sqrt{3} \approx 10 \times 1.732 \approx 17.32 \text{ m}\]
Therefore, the height of the tower is approximately 17.32 meters.
Solved Example: 9-1-05
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.
Solution:
To find the length of the string, we can use the relationship between the height of the kite, the angle of inclination of the string, and the length of the string itself.
Given:
- Height of the kite (\(h\)) = 60 m
- Angle of inclination (\(\theta\)) = 60°
Let the length of the string be \(L\).
In this scenario, we can use the sine function, as it relates the opposite side (the height of the kite) to the hypotenuse (the length of the string):
\[\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{L}\]
Substituting the known values:
\[\sin(60^\circ) = \frac{60}{L}\]
We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). So, we can substitute that in:
\[\frac{\sqrt{3}}{2} = \frac{60}{L}\]
Now, rearranging to solve for \(L\):
\[L = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}} = 40\sqrt{3}\]
Calculating the numerical value:
\[40\sqrt{3} \approx 40 \times 1.732 \approx 69.28 \text{ m}\]
Therefore, the length of the string is approximately 69.28 meters.
Solved Example: 9-1-06
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.
Solution:
To find the distance the boy walked towards the building, we can set up two right triangles based on the angles of elevation at the two positions.
Given:
- Height of the boy (\(h_b\)) = 1.5 m
- Height of the building (\(h_B\)) = 30 m
- Angle of elevation from the boy's eyes to the top of the building:
- First position: \(30^\circ\)
- Second position: \(60^\circ\)
Step 1: Calculate the effective height of the building from the boy's eye level
The height of the building above the boy's eyes is: \[h = h_B - h_b = 30\, \text{m} - 1.5\, \text{m} = 28.5\, \text{m}\]
Step 2: Set up equations for both positions
1. First Position (Angle = 30°):
Let \(d_1\) be the distance from the boy to the base of the building when he is at the first position. We can use the tangent function: \[\tan(30^\circ) = \frac{h}{d_1}\] Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\): \[\frac{1}{\sqrt{3}} = \frac{28.5}{d_1}\] Rearranging gives: \[d_1 = 28.5\sqrt{3}\]
2. Second Position (Angle = 60°):
Let \(d_2\) be the distance from the boy to the base of the building when he is at the second position. Using the tangent function again: \[\tan(60^\circ) = \frac{h}{d_2}\] Since \(\tan(60^\circ) = \sqrt{3}\): \[\sqrt{3} = \frac{28.5}{d_2}\] Rearranging gives: \[d_2 = \frac{28.5}{\sqrt{3}}\]
Step 3: Find the distance walked
The distance walked towards the building (\(d\)) is the difference between \(d_1\) and \(d_2\): \[d = d_1 - d_2\] Substituting the values: \[d = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}}\]
To simplify: \[d = 28.5\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 28.5\left(\frac{3 - 1}{\sqrt{3}}\right) = 28.5\left(\frac{2}{\sqrt{3}}\right)\] \[d = \frac{57}{\sqrt{3}} \approx 57 \times \frac{\sqrt{3}}{3} \approx 57 \times 0.577 \approx 32.9\, \text{m}\]
Final Result:
Therefore, the boy walked approximately 32.9 meters towards the building.
Solved Example: 9-1-07
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.
Solution:
To find the height of the transmission tower, we can start by analyzing the information given in the problem.
1. Define Variables:
- Let \( h \) be the height of the transmission tower.
- The total height of the transmission tower plus the building is \( 20 + h \).
2. Angles of Elevation:
- The angle of elevation to the bottom of the tower (the top of the building) is \( 45^\circ \).
- The angle of elevation to the top of the tower is \( 60^\circ \).
3. Trigonometric Relationships:
Using the tangent function, we can express the distances from the observer to the base of the building:
For the angle of elevation of \( 45^\circ \): \[\tan(45^\circ) = 1 = \frac{20}{d}\] Therefore, \[d = 20 \text{ m}\] where \( d \) is the horizontal distance from the observer to the base of the building. For the angle of elevation of \( 60^\circ \): \[\tan(60^\circ) = \sqrt{3} = \frac{20 + h}{d}\] Substituting \( d = 20 \): \[\sqrt{3} = \frac{20 + h}{20}\] Now, multiply both sides by \( 20 \): \[20\sqrt{3} = 20 + h\] Solving for \( h \): \[h = 20\sqrt{3} - 20\]
4. Calculating the Height:
Approximating \( \sqrt{3} \) (which is about \( 1.732 \)): \[h = 20(1.732) - 20\] \[h \approx 34.64 - 20\] \[h \approx 14.64 \text{ m}\] Thus, the height of the transmission tower is approximately \( 14.64 \) meters.
Solved Example: 9-1-08
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.
Solution:
Height of the Pedestal
Let the height of the pedestal be $h$ m. The height of the statue is 1.6 m. From the ground point, the angle of elevation to the top of the statue is 60°:
Using $\tan(60°) = \frac{1.6 + h}{d}$, we get $d = \frac{1.6 + h}{\sqrt{3}}$.
For the pedestal, angle of elevation is 45°:
Using $\tan(45°) = \frac{h}{d}$, we have $d = h$.
Setting both expressions for $d$ equal: $\frac{1.6 + h}{\sqrt{3}} = h$.
Solving gives $h = \frac{1.6\sqrt{3}}{\sqrt{3} - 1}$.
Solved Example: 9-1-09
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:
To solve this problem, we can use trigonometry. Let \( h \) be the height of the building.
1. From the foot of the tower:
- The angle of elevation to the top of the building is \( 30^\circ \).
- The height of the building is \( h \).
- The horizontal distance from the tower to the building is \( d \).
Using the tangent function:
\[\tan(30^\circ) = \frac{h}{d}\]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[\frac{1}{\sqrt{3}} = \frac{h}{d} \quad \Rightarrow \quad h = \frac{d}{\sqrt{3}} \quad (1)\]
2. From the foot of the building:
- The angle of elevation to the top of the tower (which is 50 m high) is \( 60^\circ \).
- The horizontal distance remains \( d \).
Using the tangent function again:
\[\tan(60^\circ) = \frac{50}{d}\]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[\sqrt{3} = \frac{50}{d} \quad \Rightarrow \quad d = \frac{50}{\sqrt{3}} \quad (2)\]
3. Substituting equation (2) into equation (1):
\[h = \frac{d}{\sqrt{3}} = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}} = \frac{50}{3}\]
Therefore, the height of the building is:
\[h = \frac{50}{3} \approx 16.67 \text{ m}\]
So, the height of the building is approximately 16.67 meters.
Solved Example: 9-1-10
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.
Solution:
Let's denote the height of the poles as \( h \). Let \( A \) and \( B \) be the positions of the poles, and let \( P \) be the point on the road between them. The distance from point \( P \) to pole \( A \) will be \( x \) meters, and thus the distance from point \( P \) to pole \( B \) will be \( 80 - x \) meters (since the road is 80 meters wide).
Step 1: Set up the equations using trigonometry
1. For pole \( A \) (angle of elevation = \( 60^\circ \)): \[\tan(60^\circ) = \frac{h}{x}\] Since \( \tan(60^\circ) = \sqrt{3} \): \[\sqrt{3} = \frac{h}{x} \quad \Rightarrow \quad h = \sqrt{3} x \quad (1)\]
2. For pole \( B \) (angle of elevation = \( 30^\circ \)):
\[\tan(30^\circ) = \frac{h}{80 - x}\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[\frac{1}{\sqrt{3}} = \frac{h}{80 - x} \quad \Rightarrow \quad h = \frac{80 - x}{\sqrt{3}} \quad (2)\]
Step 2: Set the equations equal to each other
From equations (1) and (2):
\[\sqrt{3} x = \frac{80 - x}{\sqrt{3}}\] Step 3: Solve for \( x \)
Cross-multiply: \[3x = 80 - x\] Combine like terms: \[3x + x = 80 \quad \Rightarrow \quad 4x = 80 \quad \Rightarrow \quad x = 20 \text{ m}\]
Step 4: Find the height \( h \)
Now, substitute \( x \) back into equation (1) to find \( h \):
\[h = \sqrt{3} \times 20 = 20\sqrt{3} \text{ m} \approx 34.64 \text{ m}\]
Step 5: Calculate the distance from the other pole
The distance from point \( P \) to pole \( B \) is: \[80 - x = 80 - 20 = 60 \text{ m}\]
Conclusion:
- The height of the poles is approximately \( 34.64 \) meters.
- The distance from point \( P \) to pole \( A \) is \( 20 \) meters.
- The distance from point \( P \) to pole \( B \) is \( 60 \) meters.
Solved Example: 9-1-11
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 joining 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:
Let \( h \) be the height of the tower and \( d \) be the width of the canal.
Step 1: Set up the equations:
1. From the first point (point \( A \)) directly opposite the tower:
- The angle of elevation is \( 60^\circ \).
- The horizontal distance from point \( A \) to the tower is \( d \).
Using the tangent function: \[\tan(60^\circ) = \frac{h}{d}\] Since \( \tan(60^\circ) = \sqrt{3} \): \[\sqrt{3} = \frac{h}{d} \quad \Rightarrow \quad h = \sqrt{3} d \quad (1)\]
2. From the second point (point \( B \)) which is 20 m away from point \( A \):
- The distance from point \( B \) to the tower is \( d - 20 \).
- The angle of elevation is \( 30^\circ \).
Again, using the tangent function: \[\tan(30^\circ) = \frac{h}{d - 20}\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[\frac{1}{\sqrt{3}} = \frac{h}{d - 20} \quad \Rightarrow \quad h = \frac{d - 20}{\sqrt{3}} \quad (2)\]
Step 2: Set the equations equal to each other:
From equations (1) and (2): \[\sqrt{3} d = \frac{d - 20}{\sqrt{3}}\] Step 3: Solve for \( d \) Cross-multiply: \[3d = d - 20\] Rearranging gives: \[3d - d = -20 \quad \Rightarrow \quad 2d = -20 \quad \Rightarrow \quad 2d = 20 \quad \Rightarrow \quad d = 10 \text{ m}\]Step 4: Find the height \( h \)
Substituting \( d \) back into equation (1) to find \( h \): \[h = \sqrt{3} \times 10 = 10\sqrt{3} \text{ m} \approx 17.32 \text{ m}\] Conclusion:- The height of the tower is approximately 17.32 meters.
- The width of the canal is 10 meters.
Solved Example: 9-1-12
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.
Solution:
To solve this problem, we can use trigonometry. Let's denote the height of the cable tower as \( h \).
Step 1: Define the scenario:
1. Let \( A \) be the top of the building (7 m high).
2. Let \( B \) be the foot of the cable tower.
3. Let \( C \) be the top of the cable tower.
From point \( A \):
- The angle of elevation to point \( C \) (top of the tower) is \( 60^\circ \).
- The angle of depression to point \( B \) (foot of the tower) is \( 45^\circ \).
Step 2: Set up the equations
1. Finding the horizontal distance to the tower:
- Let \( d \) be the horizontal distance from the base of the building to the foot of the cable tower.
From the angle of depression to point \( B \): \[\tan(45^\circ) = \frac{7}{d}\] Since \( \tan(45^\circ) = 1 \): \[1 = \frac{7}{d} \quad \Rightarrow \quad d = 7 \text{ m}\] 2. Finding the height of the tower \( h \): From point \( A \) to the top of the tower \( C \): \[\tan(60^\circ) = \frac{h - 7}{d}\] Since \( \tan(60^\circ) = \sqrt{3} \): \[\sqrt{3} = \frac{h - 7}{7}\] Cross-multiplying gives: \[\sqrt{3} \cdot 7 = h - 7\] \[h - 7 = 7\sqrt{3}\] \[h = 7 + 7\sqrt{3}\]
Step 3: Calculate \( h \)
Calculating \( h \): \[h = 7(1 + \sqrt{3}) \approx 7(1 + 1.732) \approx 7 \times 2.732 \approx 19.124 \text{ m}\]Conclusion:
The height of the cable tower is approximately 19.12 meters.
Solved Example: 9-1-13
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.
Solution:
To find the distance between the two ships as observed from the top of a 75 m high lighthouse, we can use trigonometry.
Step 1: Define the scenario:
- Let \( A \) be the top of the lighthouse.
- Let \( B \) be the position of the ship that is closer to the lighthouse.
- Let \( C \) be the position of the ship that is farther away from the lighthouse.
- The angles of depression to the ships are \( 45^\circ \) for ship \( B \) and \( 30^\circ \) for ship \( C \).
Step 2: Set up the equations:
1. For ship \( B \) (angle of depression = \( 45^\circ \)):
- The height of the lighthouse is \( 75 \) m.
- Let \( d_1 \) be the horizontal distance from the base of the lighthouse to ship \( B \).
Using the tangent function: \[ \tan(45^\circ) = \frac{75}{d_1}\] Since \( \tan(45^\circ) = 1 \): \[1 = \frac{75}{d_1} \quad \Rightarrow \quad d_1 = 75 \text{ m}\] 2. For ship \( C \) (angle of depression = \( 30^\circ \)):
- Let \( d_2 \) be the horizontal distance from the base of the lighthouse to ship \( C \).
Using the tangent function: \[\tan(30^\circ) = \frac{75}{d_2}\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[ \frac{1}{\sqrt{3}} = \frac{75}{d_2} \quad \Rightarrow \quad d_2 = 75\sqrt{3} \text{ m}\]
Step 3: Calculate the distance between the ships:
The distance between the two ships is: \[d_2 - d_1 = 75\sqrt{3} - 75\] Factoring out \( 75 \): \[d_2 - d_1 = 75(\sqrt{3} - 1)\]
Step 4: Calculate the numerical value:
Using \( \sqrt{3} \approx 1.732 \): \[d_2 - d_1 \approx 75(1.732 - 1) \approx 75(0.732) \approx 54.9 \text{ m}\]
Conclusion:
The distance between the two ships is approximately 54.9 meters.
Solved Example: 9-1-14
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.
Solution:
To solve the problem, we will use trigonometry to find the horizontal distances from the girl to the balloon at both angles of elevation and then calculate the distance travelled by the balloon.
Step 1: Define the scenario:
- Let \( h_g = 1.2 \) m be the height of the girl.
- Let \( h_b = 88.2 \) m be the height of the balloon from the ground.
- The height of the balloon above the girl's eyes is \( h_b - h_g = 88.2 - 1.2 = 87 \) m.
Step 2: Calculate horizontal distances:
1. When the angle of elevation is \( 60^\circ \):
- Let \( d_1 \) be the horizontal distance from the girl to the balloon at this angle.
Using the tangent function: \[\tan(60^\circ) = \frac{87}{d_1}\] Since \( \tan(60^\circ) = \sqrt{3} \): \[\sqrt{3} = \frac{87}{d_1} \quad \Rightarrow \quad d_1 = \frac{87}{\sqrt{3}} \approx 50.3 \text{ m}\] 2. When the angle of elevation is \( 30^\circ \):
- Let \( d_2 \) be the horizontal distance from the girl to the balloon at this angle.
Using the tangent function: \[\tan(30^\circ) = \frac{87}{d_2}\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[\frac{1}{\sqrt{3}} = \frac{87}{d_2} \quad \Rightarrow \quad d_2 = 87\sqrt{3} \approx 150.5 \text{ m}\]
Step 3: Calculate the distance travelled by the balloon:
The distance travelled by the balloon is: \[d_2 - d_1 = 87\sqrt{3} - \frac{87}{\sqrt{3}}\] Factoring out \( 87 \): \[d_2 - d_1 = 87\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 87\left(\frac{3 - 1}{\sqrt{3}}\right) = 87\left(\frac{2}{\sqrt{3}}\right)\] Calculating this value: \[d_2 - d_1 = \frac{174}{\sqrt{3}} \approx \frac{174 \times 1.732}{3} \approx 100.6 \text{ m}\]
Conclusion:
The distance travelled by the balloon during the interval is approximately 100.6 meters.
Solved Example: 9-1-15
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.
Solution:
To find the time taken by the car to reach the foot of the tower, we can use trigonometry to determine the distances involved.
Step 1: Define the scenario:
Let: - \( h \) be the height of the tower.
- \( d_1 \) be the horizontal distance from the car to the foot of the tower when the angle of depression is \( 30^\circ \).
- \( d_2 \) be the horizontal distance from the car to the foot of the tower when the angle of depression is \( 60^\circ \).
Step 2: Set up the equations using angles of depression:
1. When the angle of depression is \( 30^\circ \): \[\tan(30^\circ) = \frac{h}{d_1}\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[\frac{1}{\sqrt{3}} = \frac{h}{d_1} \quad \Rightarrow \quad d_1 = h \sqrt{3} \quad (1)\] 2. When the angle of depression is \( 60^\circ \): \[\tan(60^\circ) = \frac{h}{d_2}\] Since \( \tan(60^\circ) = \sqrt{3} \): \[\sqrt{3} = \frac{h}{d_2} \quad \Rightarrow \quad d_2 = \frac{h}{\sqrt{3}} \quad (2)\]
Step 3: Relate the distances:
The car moves towards the tower over a time interval of 6 seconds, so the distance it travels during this time is: $d_1 - d_2$.
Substituting equations (1) and (2): \[d_1 - d_2 = h \sqrt{3} - \frac{h}{\sqrt{3}} = h \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = h \left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}\]
Step 4: Calculate the speed of the car:
The speed of the car, \( v \), can be calculated as: \[v = \frac{\text{Distance}}{\text{Time}} = \frac{d_1 - d_2}{6} = \frac{\frac{2h}{\sqrt{3}}}{6} = \frac{h}{3\sqrt{3}} \quad (3)\]