Question

The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/hr, find the constant height at which the jet is flying.

OR

At the foot of a mountain the elevation of its summit is 45°. After ascending 1000 m toward the mountain up a slope of 30o inclination the elevation is found to be 60o. Find the height of the mountain.

Answer 1

HELLO DEAR,

let CD = h


given that:-

Speed = 720km/h ,

Time = 15s


Distance = 720 × 15/60×60 = 3km = 3000m

⇒BC = DE = 3000m


IN∆ ACD


tan30° = CD/(DE + AE)


⇒1/√3 = h/(3 + AE)


⇒√3h = (3 + AE)

⇒h = (3 + AE)/√3

⇒AE = (√3h - 3)---------(1)


NOW,

IN ∆ AEB

tan60° = BE/AE

⇒√3 = h/AE

⇒AE = h/√3------(2)

From-----(1) and---(2)

we get,

h/√3 = √3h - 3

⇒h =3h - 3√3

⇒3h - h = 3√3

⇒2h = 3×1.732

⇒h = 5.196/2

⇒h = 2.598km


∴ the height of the jet fighter is 2.598 km .



Now second answeer,



Let point A be the position of summit of the mountain and B being its foot.

Let C be the original position of observer and D, the final position after ascending 1000 metres.

Let DN and DM be perpendiculars to BC and AB respectively.


CD = 1000M = AD
∠DCN = 30°
∠ADM = 60°

IN∆DCN

Sin30° = DN/CD

⇒1/2 = DN/1000

⇒DN = 500m



IN∆ ADM,

sin60° = AM/AD

⇒√3/2 = AM/1000

⇒AM = 500√3m

⇒AM = 500×1.732 = 866m

Total height = AB = AM + BM = DN + AM = 500 + 866

⇒1366m = 1.366km

Hence, height of the mountain is 1.366 km.


I HOPE ITS HELP YOU DEAR,
THANKS