NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.4Question 14
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.4
Question. 14
From the top of a tower of height \(h\), the angles of depression of two objects in line with the foot are \(\alpha\) and \(\beta\) (with \(\beta>\alpha\)). Find the distance between the objects.
Answer:
Distance \(=h(\cot\alpha-\cot\beta)\).
Detailed Answer with Explanation:
Step 1: Draw a vertical tower of height \(h\). Let the top of the tower be point A and the bottom (foot) be point O.
Step 2: Two objects are lying on the ground in a straight line from O. Let the nearer object be at point P and the farther one at point Q.
Step 3: From the top of the tower (point A), the line of sight to P makes angle of depression \(\beta\) and the line of sight to Q makes angle of depression \(\alpha\).
Step 4: In right-angled triangle AOP:
\(\tan\beta = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{AO}{OP} = \dfrac{h}{x}\).
So, \(x = h \cot\beta\).
Step 5: In right-angled triangle AOQ:
\(\tan\alpha = \dfrac{AO}{OQ} = \dfrac{h}{y}\).
So, \(y = h \cot\alpha\).
Step 6: The distance between the two objects is:
\(PQ = y - x = h \cot\alpha - h \cot\beta\).
Final Answer: Distance \(= h(\cot\alpha - \cot\beta)\).