NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.4
Question 13

Question. 13

Two towers stand on the same plane. From the foot of the second, the angle of elevation of the top of the first (height \(30\,\text{m}\)) is \(60^\circ\). From the foot of the first, the angle of elevation of the top of the second is \(30^\circ\). Find the distance between the towers and the height of the second.

Answer:

Distance \(=10\sqrt3\,\text{m}\); height of second tower \(=10\,\text{m}.\)

Detailed Answer with Explanation:

Step 1: Understand the situation

There are two towers standing on the same horizontal ground.

  • Height of the first tower = \(30\,\text{m}\)
  • Let the height of the second tower be \(H\,\text{m}\) (unknown)
  • Let the horizontal distance between the towers be \(D\,\text{m}\)

Step 2: Use the first angle of elevation

From the foot of the second tower, the top of the first tower is seen at an angle of \(60^\circ\).

So, in the right triangle formed:

\[ \tan(60^\circ) = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{30}{D} \]

We know \(\tan(60^\circ) = \sqrt{3}\).

So, \(\sqrt{3} = \dfrac{30}{D}\).

Cross multiply: \(D = \dfrac{30}{\sqrt{3}}\).

Simplify: \(D = 10\sqrt{3}\,\text{m}\).

Step 3: Use the second angle of elevation

From the foot of the first tower, the top of the second tower is seen at an angle of \(30^\circ\).

So, in the right triangle formed:

\[ \tan(30^\circ) = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{H}{D} \]

We know \(\tan(30^\circ) = \dfrac{1}{\sqrt{3}}\).

So, \(\dfrac{1}{\sqrt{3}} = \dfrac{H}{10\sqrt{3}}\).

Cross multiply: \(H = \dfrac{10\sqrt{3}}{\sqrt{3}}\).

Simplify: \(H = 10\,\text{m}\).

Final Answer:

  • Distance between the two towers = \(10\sqrt{3}\,\text{m}\)
  • Height of the second tower = \(10\,\text{m}\)
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