Question

By observing from the top of a multi-storeyed building, the angle of depression of the top and bottom of a 4 m high building are \( 30^\circ \) and \( 45^\circ \) respectively. Find the height of the multi-storeyed building and the distance between both the buildings.

Hint:

When solving problems involving angles of depression, use the tangent function to relate the height and distance.

Answer 1

Let the height of the multi-storeyed building be \( h \) metres and the distance between the two buildings be \( d \) metres. We are given: - The height of the second building is 4 m. - The angle of depression to the top of the building is \( 30^\circ \). - The angle of depression to the bottom of the building is \( 45^\circ \). We will use the tangent of the angle to set up equations for the two triangles formed by the line of sight from the top of the multi-storeyed building. Step 1: Triangle for the top of the building
In the first triangle, using the angle of depression of \( 30^\circ \), we have: \[ \tan(30^\circ) = \frac{h - 4}{d}. \] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we get: \[ \frac{1}{\sqrt{3}} = \frac{h - 4}{d} \quad \Rightarrow \quad d = \sqrt{3}(h - 4). \] Step 2: Triangle for the bottom of the building
In the second triangle, using the angle of depression of \( 45^\circ \), we have: \[ \tan(45^\circ) = \frac{h}{d}. \] Since \( \tan(45^\circ) = 1 \), we get: \[ 1 = \frac{h}{d} \quad \Rightarrow \quad d = h. \] Step 3: Solve for \( h \)
Now, substitute \( d = h \) into the equation \( d = \sqrt{3}(h - 4) \): \[ h = \sqrt{3}(h - 4). \] Expanding the right side: \[ h = \sqrt{3} h - 4\sqrt{3}. \] Now, collect the \( h \)-terms on one side: \[ h - \sqrt{3} h = -4\sqrt{3}. \] Factor out \( h \): \[ h(1 - \sqrt{3}) = -4\sqrt{3}. \] Solve for \( h \): \[ h = \frac{-4\sqrt{3}}{1 - \sqrt{3}}. \] Now, rationalize the denominator by multiplying the numerator and denominator by \( 1 + \sqrt{3} \): \[ h = \frac{-4\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{-4\sqrt{3}(1 + \sqrt{3})}{1 - 3} = \frac{-4\sqrt{3}(1 + \sqrt{3})}{-2}. \] Simplifying: \[ h = 2\sqrt{3}(1 + \sqrt{3}). \] Now, evaluate the value of \( \sqrt{3} \approx 1.732 \): \[ h = 2 \times 1.732 \times (1 + 1.732) \approx 2 \times 1.732 \times 2.732 \approx 9.464 \, \text{m}. \] Step 4: Calculate the distance \( d \)
Since \( d = h \), we conclude that: \[ d \approx 9.464 \, \text{m}. \]
Conclusion:
The height of the multi-storeyed building is approximately \( 9.46 \, \text{m} \), and the distance between the two buildings is also approximately \( 9.46 \, \text{m} \).