Question

dn = 18 × 1.5 = 27 n.
3u × 6 0
√(40² + 27²) = √2

b. what is the bearing to the nearest degree, from the first ship to the second?

Explanation:

Step1: Identify the triangle sides

Assume from the diagram (right - angled triangle) we have two legs, let's say one leg \(a = 27\) (from \(18\times1.5 = 27\)) and the other leg \(b = 40\). We need to find the angle \(\theta\) (related to the bearing) using trigonometry. Let's assume we want to find the angle whose opposite side is \(27\) and adjacent side is \(40\) (or vice - versa depending on the bearing context). Let's use the tangent function: \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). If we consider the angle in the right - triangle, \(\tan\theta=\frac{27}{40}\) (assuming the vertical leg is \(27\) and horizontal leg is \(40\)).

Step2: Calculate the angle

First, calculate \(\tan\theta=\frac{27}{40}=0.675\). Then, \(\theta=\arctan(0.675)\). Using a calculator, \(\arctan(0.675)\approx34.09^{\circ}\). Now, depending on the bearing system (e.g., if we are measuring from the north or east direction), we need to adjust. If we assume the horizontal leg is along the east - west direction and the vertical leg is along the north - south direction, and we want the bearing from the first ship to the second, if the angle we found is with respect to the east (adjacent side is east - ward and opposite is north - ward), the bearing could be \(90^{\circ}- 34.09^{\circ}=55.91^{\circ}\) or other adjustments. Wait, maybe I mixed up the opposite and adjacent. Let's re - consider: if the vertical distance (north - south) is \(27\) and horizontal distance (east - west) is \(40\), and we want the angle from the east axis to the line connecting the two ships, \(\tan\theta=\frac{27}{40}\), \(\theta=\arctan(\frac{27}{40})\approx34^{\circ}\). But if we are finding the bearing from the first ship to the second, say the first ship is at the origin, the second ship is at \((40,27)\) (east \(40\), north \(27\)). The bearing is measured from the north or south? Wait, maybe the correct approach is: if we have a right - triangle with legs \(x = 40\) (east) and \(y = 27\) (north), the angle \(\alpha\) from the north axis would have \(\tan\alpha=\frac{40}{27}\approx1.481\), \(\alpha=\arctan(1.481)\approx56^{\circ}\). Wait, maybe I made a mistake in the opposite and adjacent. Let's start over. Let's assume that the two ships' positions form a right - triangle where one ship moves \(40\) units in one direction (say east) and \(27\) units in another (say north). To find the bearing from the first ship to the second, we can use the right - triangle. Let's use the arctangent function. If we consider the angle \(\theta\) such that \(\tan\theta=\frac{40}{27}\) (if \(40\) is the east - ward distance and \(27\) is the north - ward distance, and we want the angle from the north direction towards the east). Then \(\theta=\arctan(\frac{40}{27})\). Calculate \(\frac{40}{27}\approx1.481\), \(\arctan(1.481)\approx56^{\circ}\). So the bearing (if measured from the north, towards the east) would be \(N56^{\circ}E\) or if we are measuring from the east, it would be different. But based on the calculation of \(\arctan(\frac{40}{27})\approx56^{\circ}\) (rounded to the nearest degree).

Answer:

The bearing is approximately \(\boldsymbol{56^{\circ}}\) (the specific bearing notation may vary depending on the coordinate system, but the angle value is approximately \(56^{\circ}\))