# Ex.9.1 Q4 Some Applications of Trigonometry Solution - NCERT Maths Class 10

## Question

The angle of elevation of the top of a tower from a point on the ground, which is \(30\,\rm{m}\) away from the foot of the tower, is \(30^\circ\). Find the height of the tower.

## Text Solution

**What is Known?**

(i) Angle of elevation of the top of the tower from a point on ground is \(30^\circ\)

(ii) Distance between the foot of the tower to the point on the ground is \(30\,\rm{m.}\)

**What is Unknown?**

Height of the tower,

**Reasoning:**

Let us consider the height of the tower as \(AB,\) distance between the foot of tower to the point on ground as \(BC .\)

In \(\Delta ABC\),

Trigonometric ratio involving \(AB, BC\) and \(\angle C\) is \(\tan \theta\).

**Steps:**

In \(\Delta ABC\),

\[\begin{align}\tan \,C &= \frac{{AB}}{{BC}}\\\tan 30^\circ &= \frac{{AB}}{{30}}\\\frac{1}{{\sqrt 3 }} &= \frac{{AB}}{{30}}\\AB &= \frac{{30}}{{\sqrt 3 }}\\&= \frac{{30}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}\\

&= \frac{{30\sqrt 3 }}{3}\\&= 10\sqrt 3 \end{align}\]

Height of tower \(AB\) \(=10 \sqrt{3} \mathrm{m}\)