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

Question 17

Step 1: Draw the figure.

Imagine a window at height \(h\) from the ground. From this window:

• You look up at the top of the other house. This is the angle of elevation \(\alpha\).
• You look down at the bottom of the other house. This is the angle of depression \(\beta\).

Let the height of the other house be \(H\) and the horizontal distance between the two houses be \(d\).

Step 2: Use right-angled triangles and trigonometry.

• From the triangle formed at the top of the house:
  \(\tan\alpha = \dfrac{\text{vertical difference}}{\text{horizontal distance}} = \dfrac{H - h}{d}.\)
• From the triangle formed at the bottom of the house:
  \(\tan\beta = \dfrac{\text{vertical difference}}{\text{horizontal distance}} = \dfrac{h}{d}.\)

Step 3: Express \(d\) in terms of \(h\) and \(\beta\).

From \(\tan\beta = h/d\), rearrange to get:
\(d = \dfrac{h}{\tan\beta}.\)

Step 4: Substitute \(d\) into the first equation.

From \(\tan\alpha = (H - h)/d\):
\(H - h = d \cdot \tan\alpha.\)

Substitute \(d = h/\tan\beta\):
\(H - h = \dfrac{h}{\tan\beta} \cdot \tan\alpha.\)

Step 5: Simplify.

\(H - h = h \cdot \tan\alpha \cdot \cot\beta.\)

So, \(H = h + h \tan\alpha \cot\beta.\)

Final Result:
\(H = h \big(1 + \tan\alpha\,\cot\beta\big).\)