Question

a ship leaves port on a bearing of 43.0° and travels 14.4 mi. the ship then turns due east and travels 6.5 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 14.4 mi at a bearing of \(43.0^\circ\), then travels 6.5 mi due east. We can break the first leg into north and east components.
The north component (\(y_1\)) of the first leg: \(y_1 = 14.4\sin(43.0^\circ)\)
The east component (\(x_1\)) of the first leg: \(x_1 = 14.4\cos(43.0^\circ)\)

Step2: Calculate the total east component

The second leg is due east, so the total east component (\(x\)) is \(x = x_1 + 6.5=14.4\cos(43.0^\circ)+ 6.5\)
The north component (\(y\)) remains \(y = 14.4\sin(43.0^\circ)\)

Step3: Use the Pythagorean theorem to find the distance from port

The distance \(d\) from the port is \(\sqrt{x^{2}+y^{2}}\)
First, calculate \(x_1 = 14.4\cos(43.0^\circ)\approx14.4\times0.7314\approx10.532\)
\(x = 10.532 + 6.5 = 17.032\)
\(y = 14.4\sin(43.0^\circ)\approx14.4\times0.6820\approx9.821\)
Then \(d=\sqrt{(17.032)^{2}+(9.821)^{2}}=\sqrt{290.09 + 96.45}=\sqrt{386.54}\approx19.66\approx19.7\)

Answer:

\(19.7\)