Question

A 1.4m tall boy spots a balloon moving with the wind in a horizontal line at height of 91.4m from the ground. The angle of elevation of the balloon from the eye of the boy is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during that interval. Leave your answer nearest to metre.

Answer 1

Answer:

103.9 m

Step-by-step explanation:

Hi,

Let AD be the height of the boy

Given AD = 1.4 m

Let the balloon be at position E initially

Given BE = 91.4 m

Given ∠MDE = 60°

Let the balloon moves a distance of EF

Given that ∠FDM = 30°

Consider Δ DEM,

tan∠MDE = EM/DM

tan 60° = (BE - AD)/DM

√3 = (91.4 - 1.4)/DM

DM = 90/√3 = 30√3 m

Consider Δ DFN,

tan ∠FDN = FN/DN

tan 30° = EM/(DM + EF)

1/√3 = (BE - AD)/(DM + EF)

1/√3 = (91.4 - 1.4)/(DM + EF)

DM + EF = 90√3

EF = 90√3 - DM

= 90√3 - 30√3

= 60√3

≈ 103.9 m

Hence the distance travelled by the balloon is 103.9 m

Hope, it helps !