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

Given :-

Height of the building = 30 m

Height of the boy = 1.5 m

Inclination of the initial stand = 30°

Inclination to the other point = 60°

To Find :-

The distance boy walked towards the building.

Solution :-

(Refer to the attachment provided)

Given that,

Height of the building = 30 m

Height of the boy = 1.5 m

Inclination of the initial stand = 30°

Inclination to the other point = 60°

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

We know that,

• Point Y = Inclination 30°
• Point X = Inclination 60°

According to the question,

We have to find, the distance boy walked towards the building i.e. XY

From figure,  XY = CD.

Height of the building (AZ) = 30 m

$\longrightarrow \sf AB = AZ - BZ = 30 -1.5 = 28.5$

Measure of AB is 28.5 m

In right ΔABD,

$\sf tan \: 30^{o}=\dfrac{AB}{BD}$

$\sf \dfrac{1}{\sqrt{3} } =\dfrac{28.5}{BD}$

$\sf BD = 28.5\sqrt{3} \: m$

Again, In right ΔABC,

$\sf tan \: 60^{o}=\dfrac{AB}{BC}$

$\sf \sqrt{3} =\dfrac{28.5}{BC}$

$\sf BC=\dfrac{28.5}{\sqrt{5} } =\dfrac{28.5\sqrt{3} }{3}$

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

$\sf XY = CD = BD - BC$

$=\sf \bigg(28.5 \sqrt{3} -28.5\sqrt{\dfrac{3}{3} } \bigg)$

$\sf = 28.5 \sqrt{3} \bigg( \dfrac{1-1}{3} \bigg)$

$\sf =28.5\sqrt{3}\times \dfrac{2}{3}$

$\sf = \dfrac{57}{\sqrt{3} } \: m$

Therefore, the distance boy walked towards the building is 57/√3 m.

Answer 2

refer the attached picture..