Question

On seeing from the top of a multi-storeyed building, the angles of depression of the top and the ground of an 8 metre high house are found to be 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between both the buildings.

Hint:

In angle of depression problems, draw the diagram carefully — each line of sight forms a right triangle with the ground, and use tangent ratios to relate height and distance.

Answer 1

Step 1: Let the height of the multi-storeyed building be $h$ metres, and the horizontal distance between the buildings be $x$ metres. 
Step 2: Represent the situation. 
From the top of the taller building, the angles of depression to the top and bottom of the smaller building are 30° and 45° respectively. The height of the smaller building is 8 m. 
Step 3: Form the trigonometric equations. 
For the line of sight to the top of the smaller building: \[ \tan 30° = \frac{h - 8}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 8}{x} \Rightarrow x = \sqrt{3}(h - 8) \quad \text{...(1)} \] For the line of sight to the bottom (ground) of the smaller building: \[ \tan 45° = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h \quad \text{...(2)} \] 
Step 4: Substitute equation (2) into (1). 
\[ h = \sqrt{3}(h - 8) \] 
Step 5: Simplify the equation. 
\[ h = \sqrt{3}h - 8\sqrt{3} \Rightarrow \sqrt{3}h - h = 8\sqrt{3} \Rightarrow h(\sqrt{3} - 1) = 8\sqrt{3} \Rightarrow h = \frac{8\sqrt{3}}{\sqrt{3} - 1} \] 
Step 6: Rationalize the denominator. 
\[ h = \frac{8\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{8\sqrt{3}(\sqrt{3} + 1)}{2} = 4\sqrt{3}(\sqrt{3} + 1) \] \[ h = 4(3 + \sqrt{3}) = 12 + 4\sqrt{3} \] \[ h \approx 12 + 6.928 = 18.928 \, \text{m} \] Step 7: Find the distance between both buildings. 
From equation (2): $x = h = 18.928 \, \text{m}$. 
Step 8: Conclusion. 
\[ \boxed{\text{Height of multi-storeyed building} = 12 + 4\sqrt{3} \text{ m } (\approx 18.93 \text{ m})} \] \[ \boxed{\text{Distance between buildings} = 18.93 \text{ m}} \]