Question

The angles of depression of the top and bottom of a 10 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of the multi-storeyed building.

Hint:

In angle of depression problems, the line of sight is horizontal — use \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\) and ensure both triangles share the same base distance.

Answer 1

Step 1: Let the height of the multi-storeyed building be \( h \) m and the distance between the buildings be \( x \) m.
The height of the smaller building is \( 10 \) m.
Step 2: Apply trigonometric ratios.
From the top of the taller building, angles of depression to the top and bottom of the smaller building are 30° and 45°. For the top of the smaller building: \[ \tan 30° = \frac{h - 10}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{h - 10}{x} \Rightarrow x = \sqrt{3}(h - 10) \] For the bottom of the smaller building: \[ \tan 45° = \frac{h}{x} \] \[ 1 = \frac{h}{x} \Rightarrow x = h \]
Step 3: Equate the two values of \( x \).
\[ h = \sqrt{3}(h - 10) \]
Step 4: Simplify.
\[ h = \sqrt{3}h - 10\sqrt{3} \] \[ h(\sqrt{3} - 1) = 10\sqrt{3} \] \[ h = \frac{10\sqrt{3}}{\sqrt{3} - 1} \] Step 5: Rationalize the denominator.
\[ h = \frac{10\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = 5\sqrt{3}(\sqrt{3} + 1) \] \[ h = 5(3 + \sqrt{3}) = 15 + 5\sqrt{3} \] Step 6: Approximate value.
\[ \sqrt{3} \approx 1.732 \Rightarrow h = 15 + 8.66 = 23.66 \text{ m} \] Step 7: Conclusion.
Hence, the height of the multi-storeyed building is \( \boxed{23.66\ \text{m}} \).