A man drove a car from his office \( (O) \), 120 km on a bearing of \( 135^{\circ} \), and then drove for another 100 km on a bearing of \( 048^{\circ} \) until he reached a town \( (T) \). (a) What is the distance between the man's office and the town? (b) What is the bearing of \( O \) from \( T \) ?
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To solve this problem, we can break it down using some trigonometry and vector addition. ### (a) Distance between the man's office and the town 1. Convert the bearings to standard angles: \( 135^{\circ} \) is \( 45^{\circ} \) south of east, and \( 048^{\circ} \) is \( 48^{\circ} \) north of east. 2. Using the law of cosines, we will first find the coordinates of point \( T \): - From point \( O \) to point \( A \) (after first leg): - \( A_x = 120 \cos(135^{\circ}) = -120 \frac{\sqrt{2}}{2} \approx -84.85 \) - \( A_y = 120 \sin(135^{\circ}) = 120 \frac{\sqrt{2}}{2} \approx 84.85 \) - From point \( A \) to point \( T \): - \( T_x = 100 \cos(048^{\circ}) \approx 100 \cdot 0.6691 \approx 66.91 \) - \( T_y = 100 \sin(048^{\circ}) \approx 100 \cdot 0.7431 \approx 74.31 \) - Combine coordinates: - \( T_x = -84.85 + 66.91 \approx -17.94 \) - \( T_y = 84.85 + 74.31 \approx 159.16 \) 3. Now, calculate the distance \( OT = \sqrt{(-17.94)^2 + (159.16)^2} \approx \sqrt{321.23 + 25338.7} \approx \sqrt{25659.93} \approx 160.1 \, \text{km} \). ### (b) Bearing of \( O \) from \( T \) To find the bearing from \( T \) to \( O \): 1. Calculate the angle \( \theta \) relative to the north: \( \tan(\theta) = \frac{T_x}{T_y} = \frac{-17.94}{159.16} \). 2. Calculate \( \theta \): - Get \( \theta \approx -6.73^{\circ} \) (this angle is measured clockwise from north). 3. Adjust to find the bearing: - Since bearings are measured clockwise from north: \( 0 - (-6.73) \approx 360 - 6.73 \approx 353.27^{\circ} \). Thus, the answers to the questions are: (a) The distance between the man's office and the town is approximately \( 160.1 \, \text{km} \). (b) The bearing of \( O \) from \( T \) is approximately \( 353.27^{\circ} \).