The angles of elevation of an aeroplane flying vertically

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The angles of elevation of an aeroplane flying vertically above the ground, as observed from the two consecutive stones, 1 km apart; are 45° and 60° aeroplane from the ground is :

Two consecutive kilometre stones ⇒ C and D
∠ADB = 45°; ∠ACB = 60°
CD = 1 km.
AB = height of plane = h metre
BC = x metre (let)
In ∆ABC,

tan60° =	AB
	BC

⇒ √3 =	h
	x

⇒ h = √3x metre ..... (i)
In ∆ABD

tan45° =	AB
	BD

⇒ 1 =	h
	x + 1

⇒ h = x + 1

⇒ h =	h	+ 1
	√3

[From equation (i)]

⇒ h -	h	= 1
	√3

⇒	√3h - h	= 1
	√3

⇒ (√3 - 1)h = √3

⇒ h =	√3
	√3 - 1

⇒ h =	√3(√3 + 1)
	(√3 - 1)(√3 + 1)

⇒ h =	√3(√3 + 1)
	2

h =	(3 + √3)	metre
	2