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° = \( \dfrac{AB}{BD}\)

\( \dfrac{1}{\sqrt{3}}\) = \( \dfrac{28.5}{BD}\)

BD = \( 28.5\sqrt{3}\) m

Again,

In right ΔABC,

tan 60° = \( \dfrac{AB}{BC}\)

\(\sqrt{3}\) =\( \dfrac{28.5}{BC}\)

BC = \( 28.5\sqrt{3}\) = \(\dfrac{28.5\sqrt{3}}{3} \)

Therefore, the length of BC is \( \dfrac{28.5\sqrt{3}}{3} \) m.

XY = CD = BD – BC = (\( 28.5\sqrt{3}-\dfrac{28.5\sqrt{3}}{3} \))

= \( 28.5\sqrt{3}\)(\( 1-\dfrac{1}{3}\)) = \( 28.5\sqrt{3}\) \( \times \dfrac{2}{3}\)

= \(57\sqrt{3}\) = \(19\sqrt{3}\) m.

Thus, the distance boy walked towards the building is \( 19\sqrt{3}\) m.

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