Question

Help with Q6? "A boat travels north for 6km, west for 3km, then south for 2km. What is the boat's true bearing from its starting point? Give your answer to one decimal place.

Answer 1

`=5km`

Explanation:

Lets map this question visually. We go 6km N, 3km W and 2km S, which looks like

Since North and South are opposite direction, we can subtract their distances from each other, so we get
`=6km-2km`
`=4km` North.

Our visual map now looks like this:

Where `x` is the true bearing, or displacement of the boat. Since no angles are mentioned, we can assume that the boat turns 90 degrees to travel west, which would make this map a right-angled triangle. According to Pythagoras' Theorem, `a^2+b^2=c^2`, where `c` is the hypotenuse (longest side) and `a and b` are the other two sides.

As we can see from our map, `a=3km, b=4km, and c=x`. By substituting our known values into the formula, we get
`3^2+4^2=x^2`
`9+16=x^2`
`25=x^2`
`sqrt25=x`
`5=x`.
Thus the distance of the boat from its starting point is 5km.

This is not the end of the question, however. The question asks for the true bearing, which means we must include the angle the boat has travelled. Luckily, this is a right-angled triangle, the angles of which are already known. The largest angle is opposite the hypotenuse, `90^o`, and the smallest angle is opposite the smallest side-length, which is `36.87^o`. In this case, the smallest angle is the angle the boat has travelled, so the true bearing of the boat is 5km N`36.87^o`W.

All pictures were created in Word.

I hope I helped!