Question

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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.

☞ REQUIRED GOOD ANSWER
☞ REQUIRED FULL EXPLANATION​

Answer 1

$\huge\underline\mathrm{SOLUTION:-}$

let AB = CD = EQ = 1.5 m

Let distance covered by boy be BD = x

Let BD=AC=x

Let DQ=CE=y

In fig. PE=30-1.5=28.5m

Angles are 30° and 60°

$\small\underline\mathrm{In ∆ PAE:-}$

tan 30 = PE/AE

1/√3=28.5/x+y

28.5√3=x+y _____(1)

$\small\underline\mathrm{In ∆ PCE:-}$

tan60=PE/CE

√3=28.5/y

y=28.5/√3

Putting in equation(1)

=> x + 28.5 /√3 = 28.5√4

=> x = (28.5 × 3-28.5) /√3

=> x = (85.5-28.5) /√3

=> x =57/√3

By Rationalize:-

=57/√3×√3/√3

=57√3/3

=19√3

Answer 2

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let AB = CD = EQ = 1.5 m

Let distance covered by boy be BD = x

Let BD=AC=x

Let DQ=CE=y

Angles are 30° and 60°

In∆PAE:−

tan 30 = PE/AE

1/√3=28.5/x+y

28.5√3=x+y _____(1)

In∆PCE:−

tan60=PE/CE

√3=28.5/y

y=28.5/√3

Putting in equation(1)

=> x + 28.5 /√3 = 28.5√4

=> x = (28.5 × 3-28.5) /√3

=> x = (85.5-28.5) /√3

=> x =57/√3

By Rationalize:-

=57/√3×√3/√3

=57√3/3

=19√3

Yᴏᴜʀ Rᴇǫᴜɪʀᴇᴅ Aɴsᴡᴇʀ Is :- 19√3