Two patrol boats M3 and M7 leave a port at the same time. M3 heads due West and M7 on a bearing 282 degrees. After one hour M7 has travelled 20.1 nautical miles and observe M3 in a direction due south.

Question

a. How far is M3 from M7?
b. How far has M3 travelled?

Answer 1

a. To find the distance between M3 and M7, we can use trigonometry. Let x be the distance between them, and let y be the distance that M3 has travelled (which we will solve for in part b). Then we can draw a triangle:

```
M3 (due west)
|\
| \
| \ x
| \
| \
| \
M7----O (bearing 282 degrees)
y
```

The angle at O is 90 degrees (because M7 is due south of M3), so the angle at M7 is 90 - 282 = 78 degrees. We can use the sine rule to find x:

```
sin(78) sin(90)
------- = -------
x 20.1
```

Rearranging, we get:

```
x = sin(78) * 20.1 / sin(90) = 19.6 nautical miles (rounded to one decimal place)
```

So M3 is about 19.6 nautical miles from M7.

b. To find how far M3 has travelled, we can again use trigonometry. Let z be the distance between M3 and the point where M7 first observed it (which we can call P), and let w be the distance that M3 has travelled from the port. Again, we can draw a triangle:

```
M3 (port)---P--------O (282 degrees)
w z
```

The angle at P is 90 degrees (because M7 observed M3 due south), so the angle at O is 90 - 282 = -192 degrees (counting clockwise from due north). We can use the cosine rule to find z:

```
z^2 = w^2 + 20.1^2 - 2*w*20.1*cos(-192)
```

We can simplify the cosine term using the identity cos(-x) = cos(x):

```
z^2 = w^2 + 20.1^2 + 2*w*20.1*cos(12)
```

Now we need to use the sine rule to relate the angle between w and z (which we can call theta) to the known angle at O:

```
sin(theta) sin(12)
--------- = -------
z 20.1
```

Rearranging, we get:

```
theta = arcsin(z * sin(12) / 20.1)
```

Now we can use the sine rule again to find w:

```
sin(theta) sin(78)
--------- = -------
w 19.6
```

Rearranging and substituting for theta, we get:

```
w = sin(arcsin(z * sin(12) / 20.1)) * 19.6 / sin(78)
```

We can simplify the arcsin term using the identity arcsin(sin(x)) = x (if x is between -90 and 90 degrees):

```
w = z * sin(12) / sin(78)
```

Substituting the expression for z from earlier and simplifying, we get:

```
w = (20.1^2 - 19.6^2)^0.5 * sin(12) / sin(78) = 18.7 nautical miles (rounded to one decimal place)
```

So M3 has travelled about 18.7 nautical miles.

Answer 2

A ships sails from Sydney Harbour on a bearing 050 degrees. After the ship has sailed 30 nautical miles how far is it east of Sydney? Give your answer to the nearest nautical mile.

Answer 3

We can draw a right triangle to represent the ship's path:

```
Sydney
|
| 050°
|\
| \
| \
| \
| \
| \
| \
--------x (30 nautical miles east of Sydney)
```

The angle at Sydney is 90 degrees, and the angle opposite the distance we want to find (which we can call y) is 50 degrees. We can use trigonometry to find y:

```
y = 30 * sin(50) = 23 nautical miles (rounded to the nearest nautical mile)
```

So the ship is about 23 nautical miles east of Sydney.