Question

in the adjoining figure,if the angle of elevation is 60° and the distance AB=10√3 then the height of the tower is​

Answer 1

Answer:

The height of the tower is​ BC = 30 units, option C

Step-by-step explanation:

Given : The given triangle if right angled triangle.

            Angle A = 60°

            AB = 10$\sqrt{3}$

            Height of the tower = BC = ?

By trigonometry,  tan(A) = Opposite side BC/Adjacent side AB

                            tan 60 = BC/10$\sqrt{3}$ -------------1

we know that tan 60 = $\sqrt{3}$ ----------------------------2

Substituting 2 in 1, we get

   $\sqrt{3}$ = BC/10$\sqrt{3}$

∴ BC = 10$\sqrt{3}$ x $\sqrt{3}$

         = 10 x 3

         = 30 units

The height of the tower is​ BC = 30 units