From a window 15 metres high above the ground in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are \( 30^{\circ} \) and \( 45^{\circ} \) respectively. Find the height of the opposite house. (Use \( \sqrt{3}=1.732 \) )

To solve this problem, we can use trigonometry. 1. **Finding the distance to the first house:** Let’s denote the distance from the window to the base of the opposite house as \( d \). Since the angle of depression to the foot of the house is \( 45^{\circ} \), using the tangent function: \[ \tan(45^{\circ}) = \frac{\text{height from the window to the foot of the opposite house}}{d} = \frac{15}{d} \] Since \( \tan(45^{\circ}) = 1 \), we have: \[ 1 = \frac{15}{d} \implies d = 15 \text{ m} \] 2. **Finding the height of the opposite house:** Next, we can find the height of the opposite house, denoted as \( h \). The angle of elevation to the top of the opposite house is \( 30^{\circ} \): \[ \tan(30^{\circ}) = \frac{h - 15}{d} \] We know that \( \tan(30^{\circ}) = \frac{1}{\sqrt{3}} \). Plugging in the values, we have: \[ \frac{1}{\sqrt{3}} = \frac{h - 15}{15} \] Rearranging gives us: \[ h - 15 = \frac{15}{\sqrt{3}} \implies h = 15 + \frac{15}{\sqrt{3}} = 15 + 15 \cdot \frac{1}{1.732} \] Calculating \( \frac{15}{\sqrt{3}} \approx 8.660 \): \[ h \approx 15 + 8.660 \approx 23.660 \] So, the height of the opposite house is approximately **23.66 metres**.