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

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$\tt \: Let \: height \: of \: building$

$\: \: \: \: \: \: \: \: \tt \: AB = 30 \: m$

$\tt \: \: \: \: Height \: of \: the \: boy \: = \: PR = QM = 1.5 \: m$

$\tt \: \: \: Let \: QM \: be \: the \: position \: of \: the \: boy \: after \: walking \\ \: \: \: \: \tt some \: distance \: say \: x \: m.$

$\tt \therefore \: \: \: \: \: \: AC = AB - BC = AB - PR$

$\: \: \: \: \: \: \: \: \: \tt = 30 \: m - 1.5 \: m = 28.5$

$\: \: \tt \: In \: right \: triangle \: ACQ,$

$\: \: \: \: \: \: \: \: \: \: \: \tt \: tan \: 60 \degree \: = \frac{AC}{QC}$

$\implies \: \: \: \: \tt \: \: \: \: \: \: \sqrt{3} = \frac{28.5}{QC}$

$\implies \: \: \: \: \: \: \: \: \: \: \: \tt \: \sqrt{3} \: QC = 28.5$

$\implies \: \: \: \: \: \: \tt \: \: \: \: QC = \frac{28.5}{ \sqrt{3} }m \: \: \: \: \: \: \: \: \: \: \: \: \: \:... (1)$

$\tt \: \: \: \: \: In \: right \: triangle \: ACP,$

$\: \: \: \tt \: \: tan \: 30 \degree = \frac{AC}{PC}$

$\implies \: \: \: \: \: \tt \frac{1}{ \sqrt{3} } = \frac{28.5}{PC}$

$\implies \: \: \: \: \tt \: PC = (28.5) \sqrt{3} \: m. \: \: \: \: \: \: \: \: \: \: \: ...(2)$

$\therefore \: \: \: \: \tt \: Some \: distance \: covered \: by \: the \: boy$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt = x = RM = PQ$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: = PC - QC$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \tt = (28.5) \sqrt{3} - \frac{28.5}{ \sqrt{3} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: (by \: (2)) \: \: \: \: \: \: \: \: (by \: (1))$

$\: \: \: \: \: \: \: \: \: \: \: \: \tt \: = 28.5 \bigg( \sqrt{3} - \frac{1}{ \sqrt{3} } \bigg) = 28.5 \bigg( \frac{\sqrt{3} \sqrt{3} - 1}{ \sqrt{3} } \bigg)$

$\: \: \: \: \: \: \: \tt = \: \frac{28.5(3 - 1)}{ \sqrt{3} } = \frac{(28.5)2}{ \sqrt{3} } = \frac{57.0}{ \sqrt{3} }$

$\: \: \: \: \: \tt = \: \: \: \frac{57 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } = \frac{57 \sqrt{3} }{3} = 19 \sqrt{3}$

$\tt \: Hence, \: the \: distance \: walked \: towards \\ \tt \: the \: building \: is \: 19 \sqrt{3} \: m.$

Answer 2

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

To Find: The distance boy walked towards the building i.e. XY

From figure,

XY = CD.

Height of the building = AZ = 30 m.

AB = AZ – BZ = 30 – 1.5 = 28.5

Measure of AB is 28.5 m

In right ΔABD,

tan 30° = AB/BD

1/√3 = 28.5/BD

BD = 28.5√3 m

Again,

In right ΔABC,

tan 60° = AB/BC

√3 = 28.5/BC

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.

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