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An airplane is observed at the same time by two anti-aircraft guns $6000 m$ distant apart to be at elevations of $30 °$ and $45 °$ respectively. Find the height of the plane and its horizontal distance from the nearer gun.

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2000 (3 + 1) m

3000 (3 + 1) m

4000 (3 + 1) m

1000 (3 + 1) m

answer is C.

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Detailed Solution

Given that an airplane is observed at the same time by two anti-aircraft guns

6000 m

distant apart to be at elevations of

30 °

and

45 °

respectively.
We have to find the height of the plane and its horizontal distance from the nearer gun.
Let the height of the airplane be h. So,
Question Image

In a right-angle triangle,

tan θ = Opposite side Adjacent side

From the right angle triangle ABC,

tan 45 ° = A B B C

⇒ 1 = A B B C

∴ A B = B C

h = x … … (A)

From the right angle triangle ABD,

tan 30 ° = A B B D

⇒ 13 = h x + 6000

h (3 -1) = 6000

S-olve further for h. So,

h = 6000(3 +1) (3 -1)(3 +1) ⇒ h= 6000(3 +1) 2 ⇒ h=3000(3 +1)

As from (1) x=h,

x = 3000(3 +1)

m
The height of the plane and its horizontal distance from the nearer gun is

3000 (3 + 1) m

.
Therefore, the correct option is 3.

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