NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.4
Question 7

Question. 7

The shadow of a tower is \(50\,\text{m}\) longer when the Sun’s elevation is \(30^\circ\) than when it is \(60^\circ\). Find the height.

Answer:

\(h=25\sqrt3\,\text{m}.\)

Detailed Answer with Explanation:

Step 1: Let the height of the tower be \(h\,\text{m}\).

Step 2: When the Sun's angle of elevation is \(\theta\), the relation between height and shadow is:

\(\tan \theta = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{h}{\text{shadow length}}\)

So, \(\text{shadow length} = \dfrac{h}{\tan \theta}\).

Step 3: For \(30^\circ\):

\(L_{30} = \dfrac{h}{\tan 30^\circ}\).

Since \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\), we get:

\(L_{30} = \dfrac{h}{1/\sqrt{3}} = h\sqrt{3}\).

Step 4: For \(60^\circ\):

\(L_{60} = \dfrac{h}{\tan 60^\circ}\).

Since \(\tan 60^\circ = \sqrt{3}\), we get:

\(L_{60} = \dfrac{h}{\sqrt{3}}\).

Step 5: According to the question, the shadow at \(30^\circ\) is 50 m longer than the shadow at \(60^\circ\):

\(L_{30} - L_{60} = 50\).

Substitute the values:

\(h\sqrt{3} - \dfrac{h}{\sqrt{3}} = 50\).

Step 6: Take LHS to a single fraction:

\(\dfrac{3h - h}{\sqrt{3}} = 50\).

\(\dfrac{2h}{\sqrt{3}} = 50\).

Step 7: Multiply both sides by \(\sqrt{3}\):

\(2h = 50\sqrt{3}\).

Step 8: Divide both sides by 2:

\(h = 25\sqrt{3}\,\text{m}\).

Final Answer: The height of the tower is \(25\sqrt{3}\,\text{m}\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 8: Introduction to Trignometry and Its Applications – Exercise 8.4 | Detailed Answers