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From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:- a)149 m
- b)156 m
- c)173 m
- d)200 m
Correct answer is option 'C'. Can you explain this answer?
From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:
a)
149 m
b)
156 m
c)
173 m
d)
200 m
|
Anjana Joshi answered |
From a point P on a level ground, the angle of elevation of the top tower is 30 degrees.
Practice Quiz or MCQ (Multiple Choice Questions) with solution are available for Practice, which would help you prepare for "Height and Distance" under Logical Reasoning and Data Interpretative. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.Q.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:- a)173 m
- b)200 m
- c)273 m
- d)300 m
Correct answer is option 'C'. Can you explain this answer?
Practice Quiz or MCQ (Multiple Choice Questions) with solution are available for Practice, which would help you prepare for "Height and Distance" under Logical Reasoning and Data Interpretative. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.
Q.
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
a)
173 m
b)
200 m
c)
273 m
d)
300 m
|
Nandini Singh answered |
Let AB be the lighthouse and C and D be the positions of the ships.
A man standing on the terrace of a building watches a car speeding towards him. If at that particular instant the car is 200 m away from the building makes an angle of depression of 60° with the man’s eye and after 8 seconds the angle of depression is 30°, what is the speed of the car?- a)15 m/s
- b)25 m/s
- c)16.67 m/s
- d)Cannot be determined
Correct answer is option 'C'. Can you explain this answer?
A man standing on the terrace of a building watches a car speeding towards him. If at that particular instant the car is 200 m away from the building makes an angle of depression of 60° with the man’s eye and after 8 seconds the angle of depression is 30°, what is the speed of the car?
a)
15 m/s
b)
25 m/s
c)
16.67 m/s
d)
Cannot be determined
|
Nilesh Banerjee answered |
Understanding the Problem
The problem involves a man on a terrace observing a car. We need to determine the speed of the car based on angles of depression at two different times.
Given Data
- Initial distance from the car to the building: 200 m
- Initial angle of depression: 60°
- Angle of depression after 8 seconds: 30°
Calculating Heights and Distances
1. Initial Height Calculation:
- Using the angle of depression (60°):
- The height (h) of the terrace can be calculated using the tangent function:
- h = 200 * tan(60°) = 200 * √3
2. Position After 8 Seconds:
- After 8 seconds, the car makes an angle of depression of 30°:
- The new horizontal distance (d) from the building can be calculated:
- h = d * tan(30°)
- d = h / tan(30°) = (200 * √3) / (1/√3) = 600 m
Calculating the Distance Traveled by the Car
- Total Distance Covered:
- Initially, the car was 200 m away and is now 600 m away.
- Distance traveled = 600 m - 200 m = 400 m
Calculating Speed of the Car
- Speed Calculation:
- Speed = Distance / Time
- Speed = 400 m / 8 s = 50 m/s
Conclusion
The calculated speed of the car is incorrect in terms of the options provided. Therefore, if the answer is indeed stated as 16.67 m/s, it suggests a different interpretation or calculation might be needed. The correct approach leads to a clear understanding of the problem's requirements and the progression of the car's movement based on the angles of depression observed.
The problem involves a man on a terrace observing a car. We need to determine the speed of the car based on angles of depression at two different times.
Given Data
- Initial distance from the car to the building: 200 m
- Initial angle of depression: 60°
- Angle of depression after 8 seconds: 30°
Calculating Heights and Distances
1. Initial Height Calculation:
- Using the angle of depression (60°):
- The height (h) of the terrace can be calculated using the tangent function:
- h = 200 * tan(60°) = 200 * √3
2. Position After 8 Seconds:
- After 8 seconds, the car makes an angle of depression of 30°:
- The new horizontal distance (d) from the building can be calculated:
- h = d * tan(30°)
- d = h / tan(30°) = (200 * √3) / (1/√3) = 600 m
Calculating the Distance Traveled by the Car
- Total Distance Covered:
- Initially, the car was 200 m away and is now 600 m away.
- Distance traveled = 600 m - 200 m = 400 m
Calculating Speed of the Car
- Speed Calculation:
- Speed = Distance / Time
- Speed = 400 m / 8 s = 50 m/s
Conclusion
The calculated speed of the car is incorrect in terms of the options provided. Therefore, if the answer is indeed stated as 16.67 m/s, it suggests a different interpretation or calculation might be needed. The correct approach leads to a clear understanding of the problem's requirements and the progression of the car's movement based on the angles of depression observed.
A ship is approaching an observation tower. If the time taken by the ship to change the angle of elevation from 30° to 45° is 10 minutes, then find the time the ship will take to cover the remaining distance and reach the observation tower assuming the ship to be travelling at a uniform speed.- a)15 minutes 20 seconds
- b)13 minutes 40 seconds
- c)16 minutes 40 seconds
- d)Cannot be determined
Correct answer is option 'B'. Can you explain this answer?
A ship is approaching an observation tower. If the time taken by the ship to change the angle of elevation from 30° to 45° is 10 minutes, then find the time the ship will take to cover the remaining distance and reach the observation tower assuming the ship to be travelling at a uniform speed.
a)
15 minutes 20 seconds
b)
13 minutes 40 seconds
c)
16 minutes 40 seconds
d)
Cannot be determined
|
Bayshore Academy answered |
Let AB be the observation tower and h be its height.
Also, let the ship be at C when the angle of elevation is 30° and at D when the angle of elevation is 45°.
The time taken by the ship to travel from C to D is 10 minutes and we need to find out the time the ship will take to reach B from D.
∴ tan 30º = AB/CB = h / CB = 1/ √3
⇒ CB = √3 x h
∴ tan 45º = AB / DB = h / DB = 1
⇒ DB = h
⇒ CD = CB – DB = (√3h – h) = h(√3– 1)
Now, as h(√3 – 1) distance is covered in 10 minutes, a distance of h is covered in = 13.66 minutes = 13 minutes 40 seconds
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?- a)4√3 units
- b)8 units
- c)12 units
- d)Data inadequate
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?
a)
4√3 units
b)
8 units
c)
12 units
d)
Data inadequate
e)
None of these
|
Kalyan Mukherjee answered |
An aeroplane when 750 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 45° and 30&de; respectively. Approximately, how many meters higher is the one than the other?- a)250(1- √3)
- b)750(3- √3)
- c)250(3- √3)
- d)275(1- √3)
Correct answer is option 'C'. Can you explain this answer?
An aeroplane when 750 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 45° and 30&de; respectively. Approximately, how many meters higher is the one than the other?
a)
250(1- √3)
b)
750(3- √3)
c)
250(3- √3)
d)
275(1- √3)
|
Ssc Cgl answered |
Let C and D be the position of the aeroplanes.
Given that CB = 900 m,∠CAB = 60°,∠DAB = 45°
From the right △ ABC,
Tan45 = CB/AB => CB = AB
From the right △ ADB,
Tan30 = DB/AB => DB = ABtan30 = CBx(1/√3) = 750/√3
CB = CD + DB
=> Required height CD = CB - DB = 750 - 750/√3 = 250(3 - √3)
Two men are inverse sides of a tower. They gauge the edge of the rise of the highest point of the tower as 30° and 45° respectively. On the off chance that the tallness of the tower is 50 m, discover the separation between the two men. (Take √3=1.732)- a)135.5m
- b)136.5 m
- c)137.5 m
- d)138.5m
Correct answer is option 'B'. Can you explain this answer?
Two men are inverse sides of a tower. They gauge the edge of the rise of the highest point of the tower as 30° and 45° respectively. On the off chance that the tallness of the tower is 50 m, discover the separation between the two men. (Take √3=1.732)
a)
135.5m
b)
136.5 m
c)
137.5 m
d)
138.5m
|
EduRev SSC CGL answered |
Let AB be the tower and let C and D be the two's positions men.
At that point ∠ACB = 30°,∠ADB = 45°and AB = 50 m
AC/AB = Cot30° = √3 => AC/50 = √3
=> AC = 50√3m
AD/AB = cot 45° = 1 => AD/50 = 1
=> AD = 50M.
Separation between the two men = CD = (AC + AD)
= (50√3 + 50) m = 50(√3 + 1)
=50(1.73 + 1)m = (50 * 2.73)m = 136.5m.
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