Problem 26
Question
The angle of depression of a ship viewed at a particular instant from the top of a \(75 \mathrm{~m}\) vertical cliff is \(30 \square\). Find the distance of the ship from the base of the cliff at this instant. The ship is sailing away from the cliff at constant speed and one minute later its angle of depression from the top of the cliff is \(20^{\square}\). Determine the speed of the ship in \(\mathrm{km} / \mathrm{h}\).
Step-by-Step Solution
Verified Answer
The speed of the ship is approximately 4.56 km/h.
1Step 1: Understanding the Scenario
We have a cliff that is 75 meters tall, and from the top, we observe a ship at an angle of depression of \(30^\circ\). Using trigonometry, we can find the horizontal distance from the base of the cliff to the ship.
2Step 2: Calculating Initial Distance with Trigonometry
To find the initial distance \(d_1\) of the ship from the base of the cliff when the angle of depression is \(30^\circ\), we use the fact that \(\tan(30^\circ) = \frac{75}{d_1}\).Rearranging, we find: \[d_1 = \frac{75}{\tan(30^\circ)} = \frac{75}{\frac{1}{\sqrt{3}}} = 75\sqrt{3} \text{ meters}.\]
3Step 3: Calculating New Distance After One Minute
When the angle of depression is \(20^\circ\), the ship has moved further away. The new distance \(d_2\) can be found using \(\tan(20^\circ) = \frac{75}{d_2}\).Rearranging, we find: \[d_2 = \frac{75}{\tan(20^\circ)} \approx \frac{75}{0.3640} \approx 206.04 \text{ meters}.\]
4Step 4: Finding Distance Traveled in One Minute
The distance the ship has traveled in one minute is the difference between \(d_2\) and \(d_1\):\[d_{\text{traveled}} = d_2 - d_1 \approx 206.04 - 75\sqrt{3} \approx 76.04 \text{ meters}.\]
5Step 5: Converting Distance Traveled to Speed
To convert the distance traveled in minutes to kilometers per hour, we start by converting meters to kilometers and minutes to hours.\[76.04 \text{ m} = 0.07604 \text{ km},\]Since this distance was covered in 1 minute, the speed \(v\) in km/h is:\[v = 0.07604 \times 60 = 4.56 \text{ km/h}.\]
6Step 6: Final Answer
Based on these calculations, the speed of the ship is approximately \(4.56\text{ km/h}\).
Key Concepts
Angle of DepressionDistance CalculationSpeed of a Moving Object
Angle of Depression
When you look down from a height at an object, the angle your line of sight makes with the horizontal line is called the "angle of depression."
This is the opposite of the angle of elevation, where you look up at something. Here, the angle of depression, as in our exercise, is given from the top of a cliff.
Understanding this angle helps us figure out how far away an object is by looking at it from a higher place, like a cliff.
This concept is widely used in various real-life applications, like finding distances for navigation or construction.
This is the opposite of the angle of elevation, where you look up at something. Here, the angle of depression, as in our exercise, is given from the top of a cliff.
- The observer stands at a higher point than the object observed.
- The angle is measured between the horizontal and the line of sight down to the object.
Understanding this angle helps us figure out how far away an object is by looking at it from a higher place, like a cliff.
This concept is widely used in various real-life applications, like finding distances for navigation or construction.
Distance Calculation
Calculating distance in trigonometry often uses the tangent ratio, especially when angles of depression or elevation are involved.
In our scenario, we want to find out how far away the ship is from the base of the cliff.
Real-world applications include aviation, where planes calculate distances using similar methods.
In our scenario, we want to find out how far away the ship is from the base of the cliff.
- The distance calculation involves forming a right triangle where the height of the cliff is one side, and the distance to the ship is the other.
- We use the formula \[ \tan(\theta) = \frac{\text{Height}}{\text{Distance}} \] where \(\theta\) is the angle of depression.
Real-world applications include aviation, where planes calculate distances using similar methods.
Speed of a Moving Object
The speed of a moving object is how fast it covers a certain distance over time.
It's an important concept not just in physics, but in real-life scenarios like transportation.
This conversion ensures the speed makes sense in everyday terms, such as in driving a car.
Understanding speed and its calculation helps in grasping concepts in motion, essential for navigation and planning in various contexts.
It's an important concept not just in physics, but in real-life scenarios like transportation.
- In our exercise, the ship moves away from the cliff, changing its angular position.
- To find the speed, we determine the difference between distances at two time points.
- Speed is then calculated using the formula: \[ \text{Speed} = \frac{\text{Distance Traveled}}{\text{Time}} \]
This conversion ensures the speed makes sense in everyday terms, such as in driving a car.
Understanding speed and its calculation helps in grasping concepts in motion, essential for navigation and planning in various contexts.
Other exercises in this chapter
Problem 24
An electricity pylon stands on horizontal ground. At a point \(80 \mathrm{~m}\) from the base of the pylon, the angle of elevation of the top of the pylon is \(
View solution Problem 25
A surveyor measures the angle of elevation of the top of a perpendicular building as \(19^{\square}\). He moves \(120 \mathrm{~m}\) nearer the building and find
View solution Problem 27
In a triangle \(\mathrm{XYZ}, \angle X=51^{\circ}, \angle Y=67\) and \(Y Z=15.2 \mathrm{~cm}\). Solve the triangle and find its area.
View solution Problem 28
Solve the triangle \(\mathrm{PQR}\) and find its area given that \(Q R=36.5 \mathrm{~mm}, P R=\) \(29.6 \mathrm{~mm}\) and \(\angle Q=36\).
View solution