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An aeroplane flying at a height of 300 m above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. What is the height of the lower plane from the ground?
- a)100√3 m
- b)100/√3 m
- c)50√3 m
- d)150(√3 + 1) m
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
An aeroplane flying at a height of 300 m above the ground passes verti...
In the given figure;
BC = 300 m and suppose DC = H m
In triangle ABC;
tan60 = BC/AC
⇒ √3 = 300/AC
⇒ AC = 300/√3 m
In triangle ADC;
tan45 = DC/AC
⇒ 1 = DC/AC
⇒ DC = AC = 300/√3
∴ H = 300/√3 = 100√3 m
∴ Height of the lower plane from the ground = 100√3 m
Most Upvoted Answer
An aeroplane flying at a height of 300 m above the ground passes verti...
To solve this problem, we can use trigonometry. Let's assume that the height of the second plane above the ground is h.
From the given information, we can form a right triangle with one side being the height of the first plane (300 m), the hypotenuse being the distance between the two planes, and the angle opposite the side being 60 degrees.
Using the definition of the tangent function, we have:
tan(60) = 300 / x
Simplifying, we get:
√3 = 300 / x
Cross-multiplying, we have:
√3 * x = 300
Dividing both sides by √3, we get:
x = 300 / √3
To find the height of the second plane, we can use the same triangle. We know that the tangent of the angle of elevation for the second plane is equal to h / x.
tan(60) = h / x
Substituting the value of x we found earlier, we have:
√3 = h / (300 / √3)
Cross-multiplying, we get:
√3 * (300 / √3) = h
Simplifying, we have:
300 = h
Therefore, the height of the second plane above the ground is 300 m.
From the given information, we can form a right triangle with one side being the height of the first plane (300 m), the hypotenuse being the distance between the two planes, and the angle opposite the side being 60 degrees.
Using the definition of the tangent function, we have:
tan(60) = 300 / x
Simplifying, we get:
√3 = 300 / x
Cross-multiplying, we have:
√3 * x = 300
Dividing both sides by √3, we get:
x = 300 / √3
To find the height of the second plane, we can use the same triangle. We know that the tangent of the angle of elevation for the second plane is equal to h / x.
tan(60) = h / x
Substituting the value of x we found earlier, we have:
√3 = h / (300 / √3)
Cross-multiplying, we get:
√3 * (300 / √3) = h
Simplifying, we have:
300 = h
Therefore, the height of the second plane above the ground is 300 m.
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An aeroplane flying at a height of 300 m above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. What is the height of the lower plane from the ground?a)100√3 mb)100/√3 mc)50√3 md)150(√3 + 1) mCorrect answer is option 'A'. Can you explain this answer?
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