Question

law of cosines a ferry transports passengers to three different ports. from port a, the ferry travels east 3.5 miles to port b. the ship then turns 75° south and travels for 6 miles to reach port c. what is the distance d from port c to port a? approximate to the nearest tenth of a mile. miles

Explanation:

Step1: Determine the angle at B

The ferry travels east from A to B, then turns \(75^\circ\) south. So the angle between AB and BC is \(180^\circ - 75^\circ= 105^\circ\)? Wait, no. Wait, east to a turn of \(75^\circ\) south. So the angle between AB (east direction) and BC (which is \(75^\circ\) south of east? Wait, no, when you go east, then turn \(75^\circ\) south, the angle between AB (vector from A to B, east) and BC (vector from B to C) is \(90^\circ + 15^\circ\)? Wait, no, let's think again. If you are moving east (along positive x-axis), then turning \(75^\circ\) south means the angle between the original direction (east) and the new direction (BC) is \(180^\circ - 75^\circ = 105^\circ\)? Wait, no, actually, when you go east, then turn \(75^\circ\) towards south, the internal angle at B in triangle ABC is \(180^\circ - 75^\circ= 105^\circ\)? Wait, no, let's draw the triangle. Point A, point B is east of A (so AB is horizontal, length 3.5). Then from B, the ferry turns \(75^\circ\) south, so the direction from B to C is \(75^\circ\) below the horizontal (east) direction. So the angle at B between AB (which is towards east, from B to A is west) and BC (from B to C, \(75^\circ\) south of east) is \(180^\circ - 75^\circ = 105^\circ\)? Wait, no, the angle inside the triangle at B: AB is from A to B (east), BC is from B to C (75 degrees south of east). So the angle between BA (which is west) and BC is \(180^\circ - 75^\circ = 105^\circ\)? Wait, maybe I should use the Law of Cosines. The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and opposite angles \(A\), \(B\), \(C\) respectively, \(c^2 = a^2 + b^2 - 2ab\cos(C)\). In triangle ABC, we have AB = 3.5 miles, BC = 6 miles, and the angle at B is \(180^\circ - 75^\circ = 105^\circ\)? Wait, no, if you go east from A to B, then turn \(75^\circ\) south, the angle between AB (east) and BC (the new path) is \(90^\circ + 15^\circ\)? Wait, no, let's consider the direction. East is 0 degrees (or 90 degrees, depending on coordinate system). If we take east as the positive x-axis, then south is negative y-axis. So turning \(75^\circ\) south from east means the direction of BC is \( -75^\circ\) (or \(360 - 75 = 285^\circ\)) from the positive x-axis. The direction of AB is \(0^\circ\) (east). So the angle between AB (vector from B to A is \(180^\circ\)) and BC (vector from B to C is \(285^\circ\)) is \(285^\circ - 180^\circ = 105^\circ\). Yes, so the angle at B in triangle ABC is \(105^\circ\). So we can apply the Law of Cosines to find AC (which is \(d\)). So \(d^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle B)\). Wait, no: Law of Cosines is \(c^2 = a^2 + b^2 - 2ab\cos(C)\), where \(C\) is the angle opposite side \(c\). In triangle ABC, side opposite angle B is AC (which is \(d\)), side AB is 3.5, side BC is 6, angle at B is \(105^\circ\). So \(d^2 = 3.5^2 + 6^2 - 2 \times 3.5 \times 6 \times \cos(105^\circ)\). Wait, no, wait: the formula is \(c^2 = a^2 + b^2 - 2ab\cos(C)\), where \(C\) is the included angle between sides \(a\) and \(b\). So if we have sides AB = 3.5, BC = 6, and included angle at B is \(105^\circ\), then the side opposite angle B is AC, so \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle B)\)? Wait, no, that's not right. Wait, the Law of Cosines is \(c^2 = a^2 + b^2 - 2ab\cos(C)\), where \(C\) is the angle between sides \(a\) and \(b\). So if angle at B is between AB and BC, then the side opposite angle B is AC, so \(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle B)\)? Wait, no, that would be if the angle was acute, but if the angle is obtuse, the cosine is negative, so it becomes \(+ 2ab|\cos(\angle B)|\). Wait, let's check the formula again. Let's define: in triangle ABC, let’s denote:

- \(AB = c = 3.5\)

- \(BC = a = 6\)

- \(AC = b = d\)

- Angle at B: \(\angle B = 105^\circ\)

Then by Law of Cosines, \(b^2 = a^2 + c^2 - 2ac\cos(\angle B)\)

So \(d^2 = 6^2 + 3.5^2 - 2 \times 6 \times 3.5 \times \cos(105^\circ)\)

First, calculate \(\cos(105^\circ)\). We know that \(105^\circ = 60^\circ + 45^\circ\), so \(\cos(105^\circ) = \cos(60^\circ + 45^\circ) = \cos60^\circ\cos45^\circ - \sin60^\circ\sin45^\circ = \frac{1}{2} \times \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4} \approx \frac{1.4142 - 2.4495}{4} \approx \frac{-1.0353}{4} \approx -0.2588\)

Now, calculate each term:

\(6^2 = 36\)

\(3.5^2 = 12.25\)

\(2 \times 6 \times 3.5 = 42\)

So \(d^2 = 36 + 12.25 - 42 \times (-0.2588)\)

Calculate \(36 + 12.25 = 48.25\)

Then, \(-42 \times (-0.2588) = 42 \times 0.2588 \approx 10.87\)

So \(d^2 \approx 48.25 + 10.87 = 59.12\)

Then \(d \approx \sqrt{59.12} \approx 7.7\) (rounded to the nearest tenth)

Wait, let's check the calculation again. Wait, maybe I made a mistake in the angle. Let's re-examine the diagram. The diagram shows a triangle with A, B, C. From A to B is 3.5 miles (east), then from B to C is 6 miles, with a \(75^\circ\) angle at B (the angle between AB and BC is \(75^\circ\) south? Wait, maybe the angle at B is \(180 - 75 = 105^\circ\), but maybe it's \(75^\circ\) as the internal angle? Wait, no, if you go east, then turn \(75^\circ\) south, the angle between the original path (AB, east) and the new path (BC) is \(75^\circ\) below the east direction, so the internal angle at B in the triangle is \(180 - 75 = 105^\circ\), which is what I used. But let's recalculate \(\cos(105^\circ)\). Alternatively, use a calculator for \(\cos(105^\circ)\). \(\cos(105^\circ) \approx \cos(180^\circ - 75^\circ) = -\cos(75^\circ) \approx -0.2588\), which matches. So then:

\(d^2 = 3.5^2 + 6^2 - 2 \times 3.5 \times 6 \times \cos(105^\circ)\)

\(= 12.25 + 36 - 42 \times (-0.2588)\)

\(= 48.25 + 10.87\)

\(= 59.12\)

\(d = \sqrt{59.12} \approx 7.7\)

Wait, but let's check with another approach. Maybe the angle is \(75^\circ\) instead of \(105^\circ\). Wait, if the ferry turns \(75^\circ\) south, maybe the angle between AB (east) and BC is \(75^\circ\), but that would be if it's turning towards south from east, making the internal angle at B \(75^\circ\). Wait, that might be the mistake. Let's visualize: A is at some point, B is east of A. Then from B, the ferry turns \(75^\circ\) towards south, so the direction from B to C is \(75^\circ\) south of east. So the angle between BA (which is west) and BC is \(180^\circ - 75^\circ = 105^\circ\), but the angle between AB (east) and BC is \(75^\circ\) south, so the internal angle at B in the triangle is \(180^\circ - 75^\circ = 105^\circ\)? Wait, no, the internal angle at B is between AB and BC. AB is from A to B (east), BC is from B to C (75 degrees south of east). So the angle between AB (vector from B to A is west) and BC (vector from B to C is 75 degrees south of east) is \(180^\circ - 75^\circ = 105^\circ\). So that's correct. So the Law of Cosines should use \(105^\circ\) as the included angle. Then the calculation gives \(d \approx 7.7\). Let's verify with a calculator:

\(3.5^2 = 12.25\)

\(6^2 = 36\)

\(2 \times 3.5 \times 6 = 42\)

\(\cos(105^\circ) \approx -0.2588190451\)

So \(d^2 = 12.25 + 36 - 42 \times (-0.2588190451)\)

\(= 48.25 + 42 \times 0.2588190451\)

\(42 \times 0.2588190451 \approx 10.87039989\)

\(48.25 + 10.87039989 \approx 59.12039989\)

\(\sqrt{59.12039989} \approx 7.689 \approx 7.7\) (rounded to the nearest tenth)

So the distance from port C to port A is approximately 7.7 miles.

Answer:

\(7.7\)