A ship leaves port and travels 21km on a bearing of 032° and then 45km on a bearing of 287°.

Question

(a) calculate the distance from the port
(b) calculate the bearing of the port from the ship

Answer 1

First, let's sketch the situation:

```
032°
|
|
21 km |____________
| 287°
|
|
|
```

(a) To find the distance from the port, we can use the Pythagorean theorem to add up the horizontal and vertical distances traveled:

```
^
21 km | 45 km
---------+--------->
|
| | |
| A | B |
| | |
v-------+-------v
```

In triangle A, we have:
- Angle C = 90° (as a right angle)
- Angle A = 90° - 32° = 58° (angles in a triangle add up to 180°)
- Side a = 21 km (given)

We can use the sine function to find side b:

sin A = b / c
sin 58° = b / c
b = c * sin 58°

In triangle B, we have:
- Angle D = 90° (as a right angle)
- Angle E = 180° - 287° = 93° (angles in a triangle add up to 180°)
- Side e = 45 km (given)

We can use the sine function to find side d:

sin E = d / f
sin 93° = d / f
d = f * sin 93°

Now, we can add up the horizontal distances (b + d) and the vertical distances (a + e) to find the total distance:

distance = √((b + d)² + (a + e)²)
distance = √((21 * sin 58° + 45 * sin 93°)² + (21 * cos 58° + 45 * cos 93°)²)
distance ≈ 55.5 km

Therefore, the ship is about 55.5 km away from the port.

(b) To find the bearing of the port from the ship, we can use the inverse tangent function to find the angle between the ship's position and the port:

```
^
21 km | 45 km
---------+--------->
|
| | θ |
| A | B |
| | |
v-------+-------v
```

In triangle A, we have:
- Angle C = 90° (as a right angle)
- Angle A = 90° - 32° = 58° (angles in a triangle add up to 180°)
- Side a = 21 km (given)
- Side b = distance from part - 45km

We can use the inverse tangent function to find angle θ:

tan θ = a / b
θ = atan(a / b)

θ = atan(21 / (55.5 - 45))
θ ≈ 50.3°

Therefore, the bearing of the port from the ship is approximately 50.3°.