Question

a ship leaves port on a bearing of 38.0° and travels 10.5 mi. the ship then turns due east and travels 6.4 mi. how far is the ship from port, and what is its bearing from port? the ship is □ mi from the port. (round to the nearest tenth of a mile as needed.)

Explanation:

Step1: Analyze the ship's movement

The ship first travels 10.5 mi on a bearing of \(38.0^\circ\), then turns due east (which is a bearing of \(90^\circ\) from north, or \(0^\circ\) in standard position if we consider east as the x - axis) and travels 6.4 mi. We can break the first leg of the journey into its north - south and east - west components.
The north - south (y - axis) component of the first leg: \(y_1 = 10.5\cos(38.0^\circ)\)
The east - west (x - axis) component of the first leg: \(x_1 = 10.5\sin(38.0^\circ)\)
After the turn, the east - west component of the second leg: \(x_2=6.4\) (since it's moving due east)
The total east - west component \(x=x_1 + x_2=10.5\sin(38.0^\circ)+6.4\)
The total north - south component \(y = 10.5\cos(38.0^\circ)\)

Step2: Calculate the components

First, calculate \(x_1 = 10.5\sin(38.0^\circ)\approx10.5\times0.6157 = 6.46485\)
Then \(x=6.46485 + 6.4=12.86485\)
\(y = 10.5\cos(38.0^\circ)\approx10.5\times0.7880 = 8.274\)

Step3: Use the distance formula

The distance \(d\) from the port (origin) is given by the Pythagorean theorem \(d=\sqrt{x^{2}+y^{2}}\)
Substitute \(x = 12.86485\) and \(y = 8.274\) into the formula:
\(d=\sqrt{(12.86485)^{2}+(8.274)^{2}}=\sqrt{165.49 + 68.459}=\sqrt{233.949}\approx15.3\)

Answer:

\(15.3\)