Question

The angle of elevation of a jet fighter from a point a on the ground is 60 degree after a flight of 10 seconds the angle of elevation changes to 30 degree if the Jets flying at a speed of 432 kilometre per hour find the constant height at which the jet is flying

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Answer 1

Let the initial position of the jet fighter's plane be B

Angle of elevation from the ground to the point B is 60°.

i,e -: ∠BAC = 60°

Let the final position of the jet fighter's plane be D

Angle of elevation from the ground to the point D is 30°.

i.e -: ∠DAE = 30°

Let BD be the distance travelled by the plane with a speed of 432 km/hr.

But BC = CD as well, hence CD is also the distance covered by the plane.

Let the distance of the plane from the point C on the ground be H, therefore:

BC = DE = H

(Check the attachment for markings)

Also, speed of the jet is 432 km/hr. Converting this to m/s we get:

(To convert speed from km/hr to m/s, multiply the speed by 5/18)

$\rm \Longrightarrow 432 \times \dfrac{5}{18} \ m/s$

$\rm \Longrightarrow 120 \ m/s$

The flight is of 10 seconds, therefore the distance covered is:

⇒ Distance of BD = speed × time

⇒ Distance of BD = 120 × 10

⇒ Distance of BD = 1200m.

Now, in ΔBAC:

$\rm \Longrightarrow tan60^\circ = \dfrac{opposite \ side}{adjacent \ side}$

$\rm \Longrightarrow tan60^\circ = \dfrac{BC}{AC}$

$\rm \Longrightarrow \sqrt{3} = \dfrac{H}{AC}$

$\rm \Longrightarrow H = AC\sqrt{3}$ → (3)

In ΔDAC,

$\rm \Longrightarrow tan30^\circ = \dfrac{Opposite \ side}{Adjacent \ Side}$

$\rm \Longrightarrow tan30^\circ = \dfrac{DE}{AE}$

$\rm \Longrightarrow \dfrac{1}{\sqrt{3}} = \dfrac{H}{AE}$

$\rm \Longrightarrow H = \dfrac{AE}{\sqrt{3}}$   →  (2)

On viewing (1) & (2), we can say that both the equations are equal to H, therefore we can say that;

$\rm \Longrightarrow AC\sqrt{3} = \dfrac{AE}{\sqrt{3}}$

$\rm \Longrightarrow AC\sqrt{3} \times \sqrt{3} = AE$

$\rm \Longrightarrow 3AC = AE$

$\rm \Longrightarrow 3(AE - CE) = AE$

$\rm \Longrightarrow 3AE - 3CE = AE$

$\rm \Longrightarrow 3AE - AE = 3CE$

Substitute CE = 1200.

$\rm \Longrightarrow 2AE = 3(1200)$

$\rm \Longrightarrow 2AE = 3600$

$\rm \Longrightarrow AE = 1800m$

Substitute the value of AE in (2)

$\rm \Longrightarrow H = \dfrac{AE}{\sqrt{3}}$

$\rm \Longrightarrow H = \dfrac{1800}{\sqrt{3}}$

Rationalize the denominator:

$\rm \Longrightarrow H = \dfrac{1800}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\rm \Longrightarrow H = \dfrac{1800\sqrt{3}}{3}$

$\rm \Longrightarrow H = 600\sqrt{3}$

$\rm \Longrightarrow H \ \bold{= 1039.23 \ m}$

∴ The height at which the jet is flying from the ground is 1039.23 meters.