A pole of length 7 m is fixed vertically on the top of a tower. The angle of elevation of the top of the pole observed from a point on the ground is $60^\circ$ and the angle of depression of the same point on the ground from the top of the tower is $45^\circ$. The height (in m) of the tower is:

$\frac{7}{2}(\sqrt{3} + 1)$

⇒ In triangle ABC

⇒ tan\({60}^\circ\) = \(\frac{AB}{BC}\)

⇒ BC\(\sqrt {3 }\) = AB    ..(1.)

⇒ In triangle DBC

⇒ tan\({45}^\circ\) = \(\frac{DB}{BC}\)

⇒ 1 = \(\frac{DB}{BC}\)

⇒ BC = DB     ..(2.)

⇒ AB = AD + DB

From equation (1.) and (2.)

⇒ DB\(\sqrt {3 }\) = AB

So,

⇒ DB\(\sqrt {3 }\) = DB + 7

⇒ DB\(\sqrt {3 }\) - DB = 7

⇒ DB(\(\sqrt {3 }\) - 1) = 7

⇒ DB = $\frac{7}{2}(\sqrt{3} + 1)$ m

Therefore, The height (in m) of the tower is $\frac{7}{2}(\sqrt{3} + 1)$ m