Question

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

Answer 1

here. is the solution .

Answer 2

$\huge\star{\blue{\underline{\underline{\tt{Explanation:}}}}}$

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Given that:-

• A 1.5 m tall boy is standing at some distance from a 30 m tall building.

• The angle of elevation from his eyes to the top of the building increases from 30° to 60 ° as he walks towards the building.

To find:-

• The distance he walked towards the building.

$\mapsto$Let the boy initially standing at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.

From figure,

XY = CD.

$\leadsto$Height of the building = AZ = 30 m.

$\leadsto$AB = AZ – BZ = 30 – 1.5 = 28.5

$\leadsto$Measure of AB is 28.5 m

In right ABD

$\leadsto$tan 30° = AB/BD

$\leadsto$1/√3 = 28.5/BD

$\leadsto$BD = 28.5√3 m

Again,

In right ΔABC,

$\leadsto$tan 60° = AB/BC

$\leadsto$√3 = 28.5/BC

$\leadsto$BC = 28.5/√3 = 28.5√3/3

Therefore, the length of BC is 28.5√3/3 m

XY = CD = BD – BC = (28.5√3 – 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.

$\huge{\green{\star}}$ Thus, the distance boy walked towards the building is 57/√3 m.