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

Question. 8

A tower of unknown height is surmounted by a vertical flag staff of height \(h\). At a point, the angles of elevation of the bottom and the top of the flag staff are \(\alpha\) and \(\beta\). Prove that the height of the tower is \(\dfrac{h\tan\alpha}{\tan\beta-\tan\alpha}\).

Answer:

\(\text{Tower height}=\dfrac{h\tan\alpha}{\tan\beta-\tan\alpha}\).

Detailed Answer with Explanation:

Step 1: Let the height of the tower be \(H\), and the horizontal distance of the observation point from the tower be \(x\).

Step 2: From the point of observation, the angle of elevation to the top of the tower only is \(\alpha\). By definition of tangent in a right-angled triangle: \[ \tan\alpha = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{H}{x} \] Therefore, \[ H = x \tan\alpha. \]

Step 3: The top of the flag staff is at height \(H + h\) (tower height + flag staff height). The angle of elevation to the top of the flag staff is \(\beta\). Again, using tangent: \[ \tan\beta = \dfrac{H + h}{x} \] So, \[ H + h = x \tan\beta. \]

Step 4: We now have two equations: (i) \(H = x \tan\alpha\) (ii) \(H + h = x \tan\beta\)

Step 5: Eliminate \(x\). From (i): \[ x = \dfrac{H}{\tan\alpha}. \] Substitute this into (ii): \[ H + h = \dfrac{H}{\tan\alpha} \cdot \tan\beta. \]

Step 6: Simplify: \[ H + h = H \cdot \dfrac{\tan\beta}{\tan\alpha}. \]

Step 7: Multiply both sides by \(\tan\alpha\): \[ (H + h) \tan\alpha = H \tan\beta. \]

Step 8: Rearranging: \[ H \tan\beta - H \tan\alpha = h \tan\alpha. \] \[ H(\tan\beta - \tan\alpha) = h \tan\alpha. \]

Step 9: Finally, divide both sides by \((\tan\beta - \tan\alpha)\): \[ H = \dfrac{h \tan\alpha}{\tan\beta - \tan\alpha}. \]

Therefore, the height of the tower is proved as required.

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