Question

A boat sails 150 km on a bearing of 042° and then 250 km due east. How far is the boat from the starting point?

Answer 1

We can solve this problem using the Law of Cosines.

First, we need to draw a diagram to visualize the problem:

```
042°
|
|
| 150 km
|
|
|
----------------------
250 km to the east
```

We can see that we have a triangle with two sides and the included angle given. We need to find the third side, which is the distance between the starting point and the ending point of the boat's journey.

Let's call this third side "d".

Using the Law of Cosines, we have:

d^2 = 150^2 + 250^2 - 2(150)(250)cos(138°)

(Note that we found the angle opposite the side we want using 180° - 42° - 90° = 48°, and then used the fact that the sum of angles in a triangle is 180° to find the remaining angle to be 138°.)

Simplifying this equation, we get:

d^2 = 90000 + 62500 - 75000(-0.6691)

d^2 = 150625 + 50218.92

d^2 = 200843.92

Taking the square root of both sides, we get:

d = 448.1 km (rounded to one decimal place)

Therefore, the boat is approximately 448.1 km away from the starting point.