Question

starting at point a, a ship sails 18.7 km on a bearing of 188°, then turns and sails 47.4 km on a bearing of 319°. find the distance of the ship from point a. the distance is □□ km. (simplify your answer. type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step 1: Find the angle between the two paths

First, we need to determine the angle between the two legs of the journey. The first bearing is \(188^\circ\) and the second is \(319^\circ\). To find the angle between them, we calculate the difference: \(319^\circ - 188^\circ = 131^\circ\)? Wait, no, actually, we need to consider the direction. Let's think about the standard position (0° is north, increasing clockwise).

The first leg: bearing \(188^\circ\) means it's \(188^\circ\) clockwise from north, so the angle from the positive x - axis (if we consider north as y - axis and east as x - axis) needs to be adjusted. Wait, maybe a better way: when the ship changes direction, the angle between the two paths. Let's find the internal angle of the triangle formed by point A, the first turn point, and the final point.

The first bearing: \(188^\circ\), so the direction is \(188 - 90=98^\circ\) south of west? Wait, maybe using the law of cosines. The two sides of the triangle are \(a = 18.7\) km and \(b = 47.4\) km, and we need to find the included angle \(C\).

To find the included angle: The first bearing is \(188^\circ\), the second is \(319^\circ\). The angle between the two paths: Let's calculate the angle between the two directions. The first direction: \(188^\circ\) (clockwise from north), the second direction: \(319^\circ\) (clockwise from north). The angle between them is \(319^\circ - 188^\circ=131^\circ\)? Wait, no, because when you turn from \(188^\circ\) to \(319^\circ\), the angle between the two paths (the angle inside the triangle) is \(180^\circ-(319 - 188)^\circ\)? Wait, no, let's draw a mental picture.

Wait, the first leg: bearing \(188^\circ\) (so 8° past south towards west, since \(180^\circ\) is south, \(188 - 180 = 8^\circ\) west of south). The second leg: bearing \(319^\circ\), which is \(360 - 319=41^\circ\) east of north (since \(319^\circ\) is 41° before \(360^\circ\) (north)). Wait, maybe a better approach: the angle between the two vectors.

The first displacement vector: from A to B, length \(18.7\) km, bearing \(188^\circ\). The second displacement vector: from B to C, length \(47.4\) km, bearing \(319^\circ\). To find the angle at B (the turn point) between AB and BC, we can calculate the angle between the two bearings.

The angle between the two bearings: \(319^\circ - 188^\circ = 131^\circ\)? Wait, no, when moving from bearing \(188^\circ\) to \(319^\circ\), the external angle is \(319 - 188 = 131^\circ\), so the internal angle of the triangle (at the turn point) is \(180 - 131=49^\circ\)? No, I think I made a mistake. Let's use the formula for the angle between two bearings.

Bearing 1: \(\theta_1 = 188^\circ\) (clockwise from north)

Bearing 2: \(\theta_2 = 319^\circ\) (clockwise from north)

The angle between the two paths (the angle between the two displacement vectors) is \(|\theta_2-\theta_1 - 180^\circ|\) if \(\theta_2>\theta_1\). Wait, \(\theta_2-\theta_1=319 - 188 = 131^\circ\), then the included angle for the law of cosines is \(180 - 131 = 49^\circ\)? No, that's not right. Wait, let's consider the direction of each leg.

First leg: bearing \(188^\circ\) means the direction is \(188^\circ\) clockwise from north. So the angle with the north - south line (y - axis) is \(188 - 180 = 8^\circ\) towards west. So the components: x - component (east - west): \(18.7\sin(8^\circ)\) (west, so negative), y - component (north - south): \(18.7\cos(8^\circ)\) (south, so negative).

Second leg: bearing \(319^\circ\) means the direction is \(360 - 319 = 41^\circ\) clockwise from north? No, \(319^\circ\) clockwise from north is \(319^\circ\), so the angle with the north - south line (y - axis) is \(360 - 319 = 41^\circ\) towards east. So x - component: \(47.4\sin(41^\circ)\) (east, positive), y - component: \(47.4\cos(41^\circ)\) (north, positive).

Wait, maybe a better way is to use the law of cosines. The two sides of the triangle are \(a = 18.7\) km and \(b = 47.4\) km, and the included angle \(C\) between them. Let's find \(C\) correctly.

When the ship sails on bearing \(188^\circ\), then turns to \(319^\circ\). The angle between the two paths: Let's calculate the angle between the two directions. The first direction: \(188^\circ\) (clockwise from north), the second direction: \(319^\circ\) (clockwise from north). The angle between them is \(319 - 188 = 131^\circ\)? Wait, no, because when you are moving along \(188^\circ\), then turn to \(319^\circ\), the angle you turn is \(319 - 188 = 131^\circ\), but the included angle in the triangle (the angle between the two sides) is \(180 - 131 = 49^\circ\)? No, I think I messed up. Let's use the law of cosines with the correct included angle.

Wait, let's consider the standard method for bearing problems. The first leg: length \(a = 18.7\), bearing \(188^\circ\). The second leg: length \(b = 47.4\), bearing \(319^\circ\).

To find the angle between the two legs (the included angle \(C\)):

The difference in bearings: \(319^\circ-188^\circ = 131^\circ\), but since the ship is moving, the angle inside the triangle is \(180^\circ - 131^\circ=49^\circ\)? No, that's incorrect. Wait, let's draw the triangle. Point A, then the ship sails to point B (18.7 km, bearing 188°), then to point C (47.4 km, bearing 319°). We need to find AC.

The angle at B: when at B, the direction from B to A is opposite to the direction from A to B. The direction from A to B is bearing \(188^\circ\), so the direction from B to A is \(188^\circ - 180^\circ=8^\circ\) (bearing \(188 - 180 = 8^\circ\) north of east? Wait, no, the reverse bearing of \(188^\circ\) is \(188^\circ - 180^\circ = 8^\circ\), so bearing \(8^\circ\) (since reverse bearing is original bearing ± 180°). Then the direction from B to C is bearing \(319^\circ\). So the angle between BA (reverse of AB) and BC is \(319^\circ - 8^\circ=311^\circ\)? No, this is getting too complicated. Let's use the law of cosines with the correct included angle.

Wait, let's calculate the angle between the two vectors. The first vector (AB) has length \(18.7\), the second vector (BC) has length \(47.4\). The angle between AB and BC:

Bearing of AB: \(188^\circ\), bearing of BC: \(319^\circ\). The angle between them is \(319^\circ - 188^\circ=131^\circ\). Yes, that's the included angle for the law of cosines. Because when you are at point B, the angle between the path from A to B and B to C is \(131^\circ\). Let's verify:

If you are moving along \(188^\circ\) (almost south - west, 8° west of south) and then turn to \(319^\circ\) (almost north - west, 41° west of north), the angle between these two directions is \(131^\circ\) (because from \(188^\circ\) (8° west of south) to \(319^\circ\) (41° west of north), the angle between south - west and north - west directions: the angle between south and north is \(180^\circ\), and the west components: 8° and 41°, so total angle is \(180-(41 - 8)=147^\circ\)? No, I'm confused. Let's use the law of cosines formula \(c^2=a^2 + b^2-2ab\cos(C)\), where \(a = 18.7\), \(b = 47.4\), and \(C\) is the included angle.

Let's assume the included angle \(C = 180-(319 - 188)=49^\circ\)? No, let's calculate the angle between the two bearings correctly.

The first bearing: \(188^\circ\) (clockwise from north), so the angle with the positive y - axis (north) is \(188^\circ\), so the angle with the negative y - axis (south) is \(188 - 180 = 8^\circ\) (towards west).

The second bearing: \(319^\circ\) (clockwise from north), so the angle with the positive y - axis (north) is \(319^\circ\), so the angle with the negative y - axis (south) is \(360 - 319 = 41^\circ\) (towards east).

So the angle between the two paths (the angle between the two sides of the triangle) is \(8^\circ+41^\circ + 90^\circ=139^\circ\)? No, this is not working. Wait, let's use the formula for the angle between two bearings.

The formula for the included angle when two bearings are given: If the first bearing is \(\theta_1\) and the second is \(\theta_2\), and \(\theta_2>\theta_1\), the included angle \(C=\theta_2-\theta_1 - 180^\circ\) if \(\theta_2-\theta_1>180^\circ\), otherwise \(C = \theta_2-\theta_1\). Wait, \(\theta_2-\theta_1=319 - 188 = 131^\circ<180^\circ\), so \(C = 131^\circ\).

Now, apply the law of cosines: \(c^2=a^2 + b^2-2ab\cos(C)\), where \(a = 18.7\), \(b = 47.4\), \(C = 131^\circ\).

Step 2: Calculate \(c\) using the law of cosines

First, calculate \(a^2=18.7^2 = 349.69\)

\(b^2 = 47.4^2=2246.76\)

\(2ab\cos(C)=2\times18.7\times47.4\times\cos(131^\circ)\)

\(\cos(131^\circ)=\cos(180 - 49)^\circ=-\cos(49^\circ)\approx - 0.6561\)

So \(2\times18.7\times47.4\times(- 0.6561)=2\times18.7\times47.4\times(-0.6561)\)

First, \(2\times18.7 = 37.4\)

\(37.4\times47.4 = 37.4\times47.4 = 1772.76\)

\(1772.76\times(-0.6561)\approx - 1163.1\)

Now, \(c^2=349.69 + 2246.76-(-1163.1)=349.69 + 2246.76 + 1163.1=3759.55\)

Then \(c=\sqrt{3759.55}\approx61.3\)

Wait, let's recalculate the cosine part:

\(\cos(131^\circ)\approx\cos(131)= - 0.656059\)

\(2ab\cos(C)=2\times18.7\times47.4\times(-0.656059)\)

\(18.7\times47.4 = 18.7\times47.4 = (20 - 1.3)\times47.4=20\times47.4-1.3\times47.4 = 948 - 61.62 = 886.38\)

\(2\times886.38 = 1772.76\)

\(1772.76\times(-0.656059)\approx1772.76\times(-0.656)\approx - 1772.76\times0.656\)

\(1772.76\times0.6 = 1063.656\), \(1772.76\times0.05 = 88.638\), \(1772.76\times0.006 = 10.63656\)

Sum: \(1063.656+88.638 + 10.63656 = 1162.93056\)

So \(2ab\cos(C)\approx - 1162.93\)

Then \(c^2=a^2 + b^2-2ab\cos(C)=349.69+2246.76 + 1162.93=349.69+2246.76=2596.45;2596.45 + 1162.93 = 3759.38\)

\(c=\sqrt{3759.38}\approx61.3\) (rounded to the nearest tenth)

Answer:

\(61.3\)