The angle of elevation of a jet plane from a point A on the ground is $60^{\circ}$. After a flight of 20 seconds at the speed of $432 km /$ hour, the angle of elevation changes to $30^{\circ} .$ If the jet plane is flying at a constant height, then its height is :
- $1800 \sqrt{3} m$
- $3600 \sqrt{3} m$
- $2400 \sqrt{3} m$
- $1200 \sqrt{3} m$
The Correct Option is D
Solution and Explanation
To find the height of the jet plane flying at a constant height, we will use trigonometry. The problem is set up with two angles of elevation, and the plane travels a certain distance in the horizontal direction.
- First, convert the speed of the jet from km/h to m/s:
\(\text{Speed} = 432 \text{ km/h} = \left( \frac{432 \times 1000}{3600} \right) \text{ m/s} = 120 \text{ m/s}\)
- Calculate the horizontal distance covered by the plane in 20 seconds at this speed:
\(\text{Distance} = 120 \text{ m/s} \times 20 \text{ s} = 2400 \text{ m}\)
- Consider the point A from which the angle of elevation is observed. Initially, the angle is \(60^{\circ}\). Let the height of the plane be \(h\) and the initial horizontal distance from point A to the point right under the plane be \(x\). From the angle of elevation:
\(\tan(60^{\circ}) = \frac{h}{x}\)
\(\sqrt{3} = \frac{h}{x}\)
\(h = \sqrt{3}x\)
- After 20 seconds, the angle changes to \(30^{\circ}\), and the horizontal distance is \(x + 2400 \text{ m}\). Using this:
\(\tan(30^{\circ}) = \frac{h}{x + 2400}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{x + 2400}\)
\(h = \frac{x + 2400}{\sqrt{3}}\)
- Since both expressions represent the height \(h\), set them equal to solve for \(x\):
\(\sqrt{3}x = \frac{x + 2400}{\sqrt{3}}\)
Multiply through by \(\sqrt{3}\): \(3x = x + 2400\)
\(2x = 2400\)
\(x = 1200 \text{ m}\)
- Substitute \(x = 1200 \text{ m}\) back into the equation for \(h\):
\(h = \sqrt{3} \times 1200 = 1200\sqrt{3} \text{ m}\)
Therefore, the height of the jet plane is the correct option: \(1200 \sqrt{3} \text{ m}\).
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Concepts Used:
Trigonometric Identities
Various trigonometric identities are as follows:
Even and Odd Functions
Cosecant and Secant are even functions, all the others are odd.
- sin (-A) = – sinA,
- cos (-A) = cos A,
- cosec (-A) = -cosec A,
- cot (-A) = -cot A,
- tan (-A) = – tan A,
- sec (-A) = sec A.
Pythagorean Identities
- sin2θ + cos2θ = 1
- 1 + tan2θ = sec2θ
- 1 + cot2θ = cosec2θ
Periodic Functions
- T-Ratios of (2π + x)
sin (2π + x) = sin x,
cos (2π + x) = cos x,
tan (2π + x) = tan x,
cosec (2π + x) = cosec x,
sec (2π + x) = sec x,
cot (2π+x)=cotx. - T-Ratios of (π -x)
sin (π–x) = sin x,
cos (π–x) = - cos x,
tan (π–x) = - tan x,
cosec (π–x) = cosec x,
sec (π–x) = - sec x,
cot (π–x) = - cot x. - T-Ratios of (π+ x)
sin (π+x) = - sin x,
cos (π+x) = - cos x,
tan (π+x) = tan x,
cosec (π+x) = - cosec x,
sec (π+x) = - sec x,
cot (π+x) = cot x. - T-Ratios of (2π – x)
sin (2π–x) = - sin x,
cos (2n–x) = cos x,
tan (2π–x) = - tan x,
cosec (2π–x) = - cosec x,
sec (2π–x) = sec x,
cot (2π-x) = - cot x
Sum and Difference Identities
- T-Ratios of (x + y)
sin (x+y) = sinx.cosy + cosx.sin y
cos (x+y) = cosx.cosy – sinx.siny - T-Ratios of (x – y)
sin (x–y) = sinx.cosy – cos.x.sin y
cos (x-y) = cosx.cosy + sinx.siny
Product of T-ratios
- 2sinx cosy = sin(x+y) + sin(x–y)
- 2cosx siny = sin(x+y) – sin(x–y)
- 2 cosx cosy = cos(x+y) + cos(x–y)
- 2sinx.siny = cos(x–y) – cos(x+y)
T-Ratios of (2x)
sin2x = 2sin x cos x
cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x)
sin 3x = 3sinx – 4sin3x
cos 3x = 4cos3x – 3cosx