A vertical pole and a vertical tower are on the same level ground in such a way that, from the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. If the height of the pole is 24 m, then find the height of the tower (in m).

96

⇒ In triangle ABC

⇒ tan \({30}^\circ\) = \(\frac{24}{BC}\)

⇒ \(\frac{1}{√3}\) = \(\frac{24}{BC}\)

⇒ BC = 24\(\sqrt {3 }\)m

Now,

⇒ In triangle ADE

⇒ tan \({60}^\circ\) = \(\frac{ED}{24√3}\)

⇒ \(\sqrt {3 }\) = \(\frac{ED}{24√3}\)

⇒ ED = 24\(\sqrt {3 }\) x \(\sqrt {3 }\) = 24 x 3 = 72m

As DC = AB

⇒ EC = ED + DC

⇒ EC = 72 + 24

⇒ EC = 96m

Therefore, the height of the tower is 96m.