Question

The angle of depration of two consecutive kilometers stone on the reason right and left of an aroplane are 60°and 45° respectively as observed from the aeroplane. find the height of aeroplane . ​

Answer 1

Answer:

h = 634 m

Let h be height of the aeroplane.

Let x1 and x2 be horizontal distance of aeroplane from two kilometer stones.

x1 = h/tanθ1

x1 = h/tan60°

x1 = h/1.732 ...(1)

x2 = h/tanθ2

x2 = h/tan45°

x2 = h/1 ...(2)

Total distance 2 consecutive kilometers is 1 km = 1000.

x1 + x2 = 1000

h/1.732 + h = 1000

1000

0.5773h + h = 1000

1.5773h = 1000

h = 1000 / 1.5773

h = 634 m

Hence, height of aeroplane is 634 m.

Answer 2

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let a be the aeroplane and AD is its height

Again, let B and C be to constitutive kilometre stone on the road on the left and right of plan A and angle of depression A and B plan a are 60 and 45 respectively

Then,

∠ABC = ∠ PBC = 45° [alternative angle ]

and ∠ACB = QAC = 60° [alternative angle]

Also, BC = 1KM

Let,

BD= x KM , Then

DC = BC - BD = (1 - x) KM

in right angled Δ ADB

$→ \tan(45) = \frac{P}{B} = \frac{AD}{BD}$

$→1 = \frac{AD}{x} \: \: \: \: \: \: \: \: \: \: \: \:[{\tan(45) = 1}]$

↪ AD = x

and in right angled Δ ADC

$→ \tan(60) = \frac{AD }{DC} = \sqrt{3} = \frac{ x }{1 - x} \: \: \: \: \: \: \: \: \: \: \: \: \: \: [: \tan(60) = \sqrt{3} ]$

• $→ \sqrt{3} - \sqrt{3x} = x$
• $→ \sqrt{3} = \sqrt{3x} + x$
• $→( \sqrt{3} + 1)x = \sqrt{3}$
• $→x = \frac{ \sqrt{3} }{ \sqrt{3 + 1} }$
• $→ \frac{ \sqrt{3} }{ \sqrt{3 + 1} } \times \frac{ \sqrt{3 - 1} }{ \sqrt{3 - 1} }$
• $→ \frac{3 - \sqrt{3} }{( \sqrt{3} ^{2} ) - (1) ^{2} }$
• $→ \frac{3 - \sqrt{3} }{2}$
• $→ \frac{3 - 1.732}{2} = 0.634 \: km$

Or,

• $→0.634 \times 1000 = 634 \: m$

[ 1 KM = 1000 M ]

Final answer is 634 m.