Let CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of the building and the tower. Elevation is 30 degree and 60 degree from the tower and the building respectively.

In right ΔBCD,

tan 60° = \(\dfrac{CD}{BC} \)

\( \sqrt{3}\) = \( \dfrac{50}{BC}\)

BC = \( \dfrac{50}{\sqrt{3}}\) …(1)

Again,

In right ΔABC,

tan 30° = \( \dfrac{AB}{BC} \)

⇒ \( \dfrac{1}{\sqrt{3}}\) = \( \dfrac{AB}{BC} \)

Use result obtained in equation (1)

AB = \( \dfrac{50}{3} \)

Thus, the height of the building is \( \dfrac{50}{3} \) m.

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