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 \( T \) ?

To solve this problem, we can use the concept of vectors and trigonometry. Let's denote the man's office as point \( O \), the town as point \( T \), and the direction from the office to the town as vector \( \vec{OT} \). Given: - Distance from office to town: 120 km - Bearing from office to town: \( 135^{\circ} \) - Distance from office to town: 100 km - Bearing from office to town: \( 048^{\circ} \) We can break down the vector \( \vec{OT} \) into two components: - Horizontal component: \( 120 \cos(135^{\circ}) \) - Vertical component: \( 120 \sin(135^{\circ}) \) Similarly, for the second part of the journey: - Horizontal component: \( 100 \cos(048^{\circ}) \) - Vertical component: \( 100 \sin(048^{\circ}) \) Now, we can calculate the total horizontal and vertical components of vector \( \vec{OT} \) by adding the components from both parts of the journey. (a) The distance between the man's office and the town is the magnitude of vector \( \vec{OT} \), which can be calculated using the Pythagorean theorem. (b) The bearing of the man's office from the town can be found by taking the inverse tangent of the ratio of the vertical component to the horizontal component of vector \( \vec{OT} \). Let's calculate the components and the distance between the office and the town. Calculate or simplify the expression \( 120 * sin(135) + 100 * sin(48) \). Calculate or simplify the expression \( \sqrt{(-183.544974)^2 + (-66.221224)^2} \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{\left(-183.544974\right)^{2}+\left(-66.221224\right)^{2}}\) - step1: Convert the expressions: \(\sqrt{\left(-\frac{91772487}{500000}\right)^{2}+\left(-66.221224\right)^{2}}\) - step2: Convert the expressions: \(\sqrt{\left(-\frac{91772487}{500000}\right)^{2}+\left(-\frac{8277653}{125000}\right)^{2}}\) - step3: Add the numbers: \(\sqrt{\frac{91772487^{2}+16\times 8277653^{2}}{500000^{2}}}\) - step4: Use the properties of radicals: \(\frac{\sqrt{91772487^{2}+16\times 8277653^{2}}}{\sqrt{500000^{2}}}\) - step5: Simplify the expression: \(\frac{\sqrt{91772487^{2}+16\times 8277653^{2}}}{500000}\) The distance between the man's office and the town is approximately 195.13 km. (b) To find the bearing of the man's office from the town, we need to calculate the angle using the inverse tangent function. Let's calculate the angle. Calculate the value by following steps: - step0: Calculate: \(\arctan\left(\frac{-66.221224}{\left(-183.544974\right)}\right)\) - step1: Remove the parentheses: \(\arctan\left(\frac{-66.221224}{-183.544974}\right)\) - step2: Divide the terms: \(\arctan\left(\frac{33110612}{91772487}\right)\) The bearing of the man's office from the town is approximately 20.2°.