A ship sailed from port A to port B on the bearing N65°E a distance of 20km sailed from port B to port C on S25°E, a distance of 240km.
a) Calculate the distance between A and C
b) Find i) Bearing of C from A ii)Bearing of A from C
As always, draw the diagram. You will see that you have a right triangle with legs of 20 and 240, so you can easily figure
(a) the hypotenuse.
(b) the angle at C has tanC = 20/240, so you can easily figure the direction of the hypotenuse. The bearings are 180° apart.
All angles are measured CW from +y-axis.
AC = AB + BC = 20km[65o] + 240km[155o],
X = 20*sin65 + 240*sin155 = 119.6 km.
Y = 20*Cos65 + 240*Cos155 = -209.1 km.
a. AC = sqrt(X^2+Y^2) = 241 km.
b. Tan A = X/Y,
A = -29.8o = 29.8o E. of S. = 150.2o CW = bearing of C from A.
150.2 + 180 = __Degrees = bearing of A from C.
To calculate the distance between points A and C, we can use the concept of vector addition. We visualize the movement of the ship as two separate displacements in the form of vectors.
a) Calculate the distance between A and C:
First, let's break down the movement from A to C into two components:
1. Movement from A to B:
This corresponds to a bearing of N65°E and a distance of 20km. Let's denote this vector as vector AB.
2. Movement from B to C:
This corresponds to a bearing of S25°E and a distance of 240km. Let's denote this vector as vector BC.
To find the resultant displacement (vector AC), we need to add these two vectors:
AC = AB + BC
To add the vectors, we can use the concept of vector components. Imagine creating a right-angled triangle with vectors AB and BC as adjacent sides. By using trigonometry, we can find the magnitude and direction of the resultant vector.
i) Finding the magnitude of AC:
Using the Pythagorean theorem, the magnitude of AC can be found as:
|AC| = sqrt((|AB|)^2 + (|BC|)^2)
|AB| = 20km (given)
|BC| = 240km (given)
|AC| = sqrt((20)^2 + (240)^2)
= sqrt(400 + 57,600)
= sqrt(58,000)
≈ 240.83km
Therefore, the distance between A and C is approximately 240.83 km.
b) Finding the bearings from A to C and from C to A:
i) Bearing of C from A:
To find the bearing of C from A, we can use trigonometry. We need to determine the angle between the line AC and the North direction.
In right-angled triangle ABC, consider the angle at B. This angle corresponding to the bearing of C from A can be found using the cosine rule:
cos(B) = (|AC|^2 + |AB|^2 - |BC|^2) / (2 * |AC| * |AB|)
Substituting the values:
cos(B) = (240.83^2 + 20^2 - 240^2) / (2 * 240.83 * 20)
= (58,000 + 400 - 57,600) / 9616.6
= 800 / 9616.6
≈ 0.08307
B = arccos(cos(B))
≈ 84.68°
Since the bearing is the direction measured clockwise from North, the bearing of C from A is N84.68°E.
ii) Bearing of A from C:
To find the bearing of A from C, we simply need to subtract the bearing of C from A (180°) to get the opposite direction.
Bearing of A from C = N84.68°E - 180°
= S95.32°E
Therefore, the bearing of A from C is S95.32°E.