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From an aeroplane above a straight road the angle of depression of two positions at a distance 20 m apart on the road are observed to be 30° and 45°. The height of the aeroplane above the ground is :
- a)10√3m
- b)10(√3-l)m
- c)10(√3+l)m
- d)20m
Correct answer is option 'C'. Can you explain this answer?
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From an aeroplane above a straight road the angle of depression of two...
Hence the height is 10(√3+l)m
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From an aeroplane above a straight road the angle of depression of two...
To solve this problem, we can use the concept of trigonometry and the properties of angles of depression.
Let's assume that the aeroplane is flying at point A above the road. The two positions observed on the road are B and C, with a distance of 20 m apart.
We can draw a right-angled triangle ABC, where angle BAC is the angle of depression from the aeroplane to point B, and angle CAB is the angle of depression from the aeroplane to point C.
Since the angle of depression is defined as the angle formed between the line of sight from the observer (in this case, the aeroplane) to the object (point B or C) and a horizontal line, we can say that angle BAC and angle CAB are both 30 degrees.
We know that the distance between points B and C is 20 m.
To find the height of the aeroplane above the road, we can use the tangent function.
Tangent of an angle is equal to the opposite side divided by the adjacent side.
In this case, the opposite side is the height of the aeroplane above the road, and the adjacent side is the distance between points B and C.
So, we have:
tan(30 degrees) = height of the aeroplane / 20 m
Using a calculator, we can find that the tangent of 30 degrees is approximately 0.5774.
Therefore,
0.5774 = height of the aeroplane / 20 m
To find the height of the aeroplane, we can rearrange the equation:
height of the aeroplane = 0.5774 * 20 m
height of the aeroplane = 11.548 m
Therefore, the height of the aeroplane above the road is approximately 11.548 meters.
Let's assume that the aeroplane is flying at point A above the road. The two positions observed on the road are B and C, with a distance of 20 m apart.
We can draw a right-angled triangle ABC, where angle BAC is the angle of depression from the aeroplane to point B, and angle CAB is the angle of depression from the aeroplane to point C.
Since the angle of depression is defined as the angle formed between the line of sight from the observer (in this case, the aeroplane) to the object (point B or C) and a horizontal line, we can say that angle BAC and angle CAB are both 30 degrees.
We know that the distance between points B and C is 20 m.
To find the height of the aeroplane above the road, we can use the tangent function.
Tangent of an angle is equal to the opposite side divided by the adjacent side.
In this case, the opposite side is the height of the aeroplane above the road, and the adjacent side is the distance between points B and C.
So, we have:
tan(30 degrees) = height of the aeroplane / 20 m
Using a calculator, we can find that the tangent of 30 degrees is approximately 0.5774.
Therefore,
0.5774 = height of the aeroplane / 20 m
To find the height of the aeroplane, we can rearrange the equation:
height of the aeroplane = 0.5774 * 20 m
height of the aeroplane = 11.548 m
Therefore, the height of the aeroplane above the road is approximately 11.548 meters.
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From an aeroplane above a straight road the angle of depression of two positions at a distance 20 m apart on the road are observed to be 30° and 45°. The height of the aeroplane above the ground is :a)10√3mb)10(√3-l)mc)10(√3+l)md)20mCorrect answer is option 'C'. Can you explain this answer?
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