Heights and Distances Practice Test 2

Question : 12 of 24

Marks: +1, -0

An aeroplane when flying at a height of 5,000 m from the ground passes vertically above another aeroplane at an instant, when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. The vertical distance between the aero planes at that instant is:

$5000(√3-1)m$
$5000(√3-√3)m$
$5000(1-1/√3 )m$
4,500 m

Validate

Solution:

∠ACB = 60°
∠DCB = 45°
AB = 5,000 meter
AD = x, meter
∴ From ΔABC

|tan 60°={AB}/{BC}

⇒√3=5000/{BC}

⇒BC=5000/√3m

From ΔDBC

\tan45°={DB}/{BC}⇒DB=BC=5000/√3

AD = AB - BD

=5000-5000/√3=5000({√3-1}/√3)m

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