Question

A 1.5m tall boy is standing at some distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increase from 30° to 60° as he walks towards the building. Find the distance he walked towards the building​

Answer 1

Step-by-step explanation:

Given :-

A 1.5m tall boy is standing at some distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increase from 30° to 60° as he walks towards the building.

To find :-

Find the distance he walked towards the building ?

Solution :-

On converting the above data into pictorial representation then

Height of the boy = AB = 1.5 m

Height of the building = GE = 30 m

AB = EF = 1.5 m

GE = GF + FE

=> 30 = GF + 1.5

=> GF = 30-1.5

=> GF = 28.5 m

The boy walks towards the building from A to D

Then the angle of elevation changes 30° to 60°

Total distance = AD+DF = AF

The boy walks towards the building =

AD = BC

angles of elevation are < GAF = 30° and

< GDF = 60°

In ∆ GAF ,

Tan30°=Opposite side to 30°/Adjacent side to 30°

=> Tan 30° = GF/AF

=> 1/√3 = 28.5/AF

On applying cross multiplication then

=> AF× 1 = 28.5×√3

=> AF = 28.5√3 ------------(1)

and

In ∆ GDF ,

Tan60°=Opposite side to 60°/Adjacent side to 60°

=> Tan 60° = GF/DF

=> √3 = 28.5/DF

On applying cross multiplication then

=> DF×√3= 28.5

=> DF = 28.5/√3 ------------(2)

We have

AD+DF = AF

AD = AF-DF

=> AD = 28.5√3-(28.5/√3)

=> AD = 28.5[√3-(1/√3)]

=> AD = 28.5[(3-1)/√3]

=>AD = 28.5×(2/√3)

=> AD = 57/√3

=> AD = (19×3)/√3

=> AD = (19×√3×√3)/√3

=> AD = 19√3 m

We know that √3 = 1.732

=> AD = 19×1.732 m

=> AD = 32.908 m

=> AD = 32.91 m

Answer:-

The boy walks towards the building for the given problem is 19√3 m or 32.91 m

Used formulae:-

• Tan A =Opposite side to A/ Adjacent side to A
• Tan 30° = 1/√3
• Tan 60° = √3
• √3 = 1.732...