Question

An aeroplane (A) when flying at a certain height from the ground passes vertically above another aeroplane (B) at an instant when the angles of the elevation of the two aeroplanes from the same point on the ground are 30° and 45° respectively and the distance between two aeroplanes is 1000 m. Then, the approximate vertical height of the aeroplane A is

Answer 1

Solution :-

From image we have :-

→ AC = Let h m.

→ CD = d m.

→ BA = 1000m.

→ ∠C = 90° .

→ ∠ADC = 30°

→ ∠BDC = 45° .

in Right ∆ACD , we have,

→ Tan30° = P/B = AC/CD

→ (1/√3) = h /d

→ d = √3*h ----------- Eqn.(1)

in Right ∆BCD, we have,

→ Tan45° = P/B = BC/CD

→ 1 = (1000 + h)/d

→ d = (1000 + h) ---------- Eqn.(2)

So, From Eqn.(1) and Eqn.(2) , we get,

→ √3h = (1000 + h)

→ √3h - h = 1000

→ h(√3 - 1) = 1000

→ h = 1000 / (√3 - 1)

Rationalize the RHS Part,

→ h = { 1000 / (√3 - 1) } * {(√3 + 1)/(√3 + 1)}

→ h = 1000(√3 + 1) / (3 - 1)

→ h = 1000(√3 + 1)/2

→ h = 500(√3 + 1)

Putting √3 = 1.73 now,

→ h = 500(1.73 + 1)

→ h = 500 * 2.73

→ h = 1365 m.

Hence, the approximate vertical height of the aeroplane A is 1365 m.