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 (b) Whn's office and the town? (b) is the bearing of \( O \) from \( T \) ?

To find the distance and bearing between the origin (Whn's office) and the town (T), we can use trigonometry and the law of cosines. First, let's represent the movements: 1. From origin \( O \) to point \( A \), a vector of \( 120 \) km at a bearing of \( 135^{\circ} \). 2. From point \( A \) to point \( T \), a vector of \( 100 \) km at a bearing of \( 048^{\circ} \). The angle between the two paths can be calculated by adjusting the bearings into standard position. - The angle of \( 135^{\circ} \) corresponds to \( 180^{\circ} - 135^{\circ} = 45^{\circ} \) to the south of east. - The angle of \( 048^{\circ} \) is \( 48^{\circ} \) to the north of east. Now, find the angle \( \theta \) between the two travels: \[ \theta = 135^{\circ} - 48^{\circ} = 87^{\circ}. \] Now we can apply the law of cosines to find the distance \( d \) as: \[ d^2 = 120^2 + 100^2 - 2 \times 120 \times 100 \times \cos(87^\circ). \] Calculating \( d \): - \( 120^2 + 100^2 = 14400 + 10000 = 24400. \) - The cosine of \( 87^\circ \) is quite small, so \( \cos(87^\circ) \approx 0.0872 \). - Thus, \( 2 \times 120 \times 100 \times \cos(87^\circ) \approx 12000 \times 0.0872 \approx 1046.4. \) Now substituting: \[ d^2 \approx 24400 - 1046.4 \approx 23353.6, \] \[ d \approx \sqrt{23353.6} \approx 152.14 \text{ km}. \] For part (b), we find the bearing of \( O \) from \( T \). To determine it, we use the known angles and the geometry of the triangle formed. Since bearing is measured clockwise from north: 1. \( O \) to \( A \) makes an angle of \( 135^\circ \) (south-east). 2. From \( A \) to \( T \) is \( 048^\circ \). The bearing from \( T \) to \( O \) can be found by adding \( 180^\circ \) to the angle between \( T \) and \( O \): \[ \text{Bearing from } T \text{ to } O = 180^\circ + 048^\circ = 228^\circ. \] Thus, (a) The distance between Whn's office and the town is approximately \( 152.14 \) km. (b) The bearing of \( O \) from \( T \) is \( 228^\circ \).