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:
Answer & explanation
Correct answer: option 4
⇒ 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