Building Height Calculation Using Angles of Depression

This problem involves calculating the height of a building based on angles of depression to two houses located on the ground.

Problem Setup

• Let the height of the building be H meters.
• Let the points on the ground directly below the top of the building be B, and the top of the building be A.
• Let the positions of the two houses be C and D. Both houses are at the same height (ground level).
• The angles of depression from A to C and D are $45^{\circ}$ and $30^{\circ}$ respectively, towards the east.
• This means the angle of elevation from C to A is $45^{\circ}$, and from D to A is $30^{\circ}$.
• Since the angles of depression are different, the houses C and D are at different distances from the base B. The house with the larger angle of depression ($45^{\circ}$) is closer.
• Let the distance from the base B to the closer house C be $x$ meters (BC = $x$).
• Let the distance from the base B to the farther house D be $y$ meters (BD = $y$).
• The two houses are 50 m apart, and they are towards the east. Assuming C is between B and D, we have $y - x = 50$.

Trigonometric Relationships

We can use the tangent function in the right-angled triangles formed:

• Triangle ABC (right-angled at B): $\tan(45^{\circ}) = \frac{AB}{BC} = \frac{H}{x}$
• Triangle ABD (right-angled at B): $\tan(30^{\circ}) = \frac{AB}{BD} = \frac{H}{y}$

Solving for Height (H)

1. From triangle ABC: Since $\tan(45^{\circ}) = 1$, we have $1 = \frac{H}{x}$. This gives us $x = H$.
2. From triangle ABD: Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$, we have $\frac{1}{\sqrt{3}} = \frac{H}{y}$. This gives us $y = H\sqrt{3}$.
3. Using the distance between houses: We know $y - x = 50$. Substitute the expressions for $x$ and $y$ in terms of $H$: $H\sqrt{3} - H = 50$
4. Factor out H: $H(\sqrt{3} - 1) = 50$
5. Solve for H: $H = \frac{50}{\sqrt{3} - 1}$
6. Rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{3} + 1)$: $H = \frac{50}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$ $H = \frac{50(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$ $H = \frac{50(\sqrt{3} + 1)}{3 - 1}$ $H = \frac{50(\sqrt{3} + 1)}{2}$
7. Simplify the expression: $H = 25(\sqrt{3} + 1)$

The height of the building is $25(\sqrt{3} + 1)$ metres.