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 \( \sqrt{3}=1.732 \) )

Let's break down the problem step by step. **Given:** - The height of the window from the ground: \( 15 \) meters. - Angle of depression to the foot of the opposite house: \( 45^\circ \). - Angle of elevation to the top of the opposite house: \( 30^\circ \). - \( \sqrt{3} = 1.732 \). **Objective:** Find the height (\( h \)) of the opposite house. **Step 1: Determine the Horizontal Distance (\( d \)) Between the Buildings** From the window, looking down at an angle of depression of \( 45^\circ \) to the foot of the opposite house: \[ \tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{15}{d} \] Since \( \tan(45^\circ) = 1 \): \[ 1 = \frac{15}{d} \implies d = 15 \text{ meters} \] **Step 2: Determine the Additional Height Viewed at an Angle of Elevation of \( 30^\circ \)** From the window, looking up at an angle of \( 30^\circ \) to the top of the opposite house: \[ \tan(30^\circ) = \frac{h - 15}{15} \] Given \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx \frac{1}{1.732} \): \[ \frac{1}{1.732} = \frac{h - 15}{15} \] Multiply both sides by \( 15 \): \[ \frac{15}{1.732} = h - 15 \] \[ h = 15 + \frac{15}{1.732} \] Calculate \( \frac{15}{1.732} \): \[ \frac{15}{1.732} \approx 8.66 \text{ meters} \] Thus, \[ h \approx 15 + 8.66 = 23.66 \text{ meters} \] **Final Answer:** The height of the opposite house is approximately **23.66 meters**.