1. We are given a triangle with points T (top of building), P, Q, R on the ground. 2. Given data: - \( \sqrt{2} = 1.41 \), \( \sqrt{3} = 1.73 \) - \( \angle QRT = 30^\circ \) - \( \angle TQP = 45^\circ \) - Length \( QR = 60 \) meters 3. The goal is to find the height of the building (segment TP). 4. Step 1: Analyze triangle QRT with \( \angle QRT = 30^\circ \) and side \( QR = 60 \). - Using triangle properties, length \( QT = QR \times \sqrt{3} = 60 \times 1.73 = 103.8 \) meters 5. Step 2: Analyze triangle TQP with \( \angle TQP = 45^\circ \). - Since \( \angle TQP = 45^\circ \) and TP is vertical, triangle TQP is right-angled and isosceles - So, \( TP = PQ \) 6. Step 3: Length PQ can be calculated from points Q and P along the base. - Using the properties from previous triangle and angles, \( PQ = QT \times \sin 45^\circ = 103.8 \times 0.707 = 73.4 \) meters 7. Step 4: Therefore, height of building \( TP = PQ = 73.4 \) meters. Answer: The building's height \( TP \) is approximately \( \boxed{73.4} \) meters.