Question

Barfield is 7km north and 8km east of Westgate. The bearing to get from Westgate to Barfield is 041.2, and Lauren sails at a bearing of 043. She stops when she is due North of Barfield. How far away is she from Barfield?

PLEASE HELP!
I am struggling with a bearings question that came up on a recent ks3 homework sheet, and having not done bearings for over a year, I cannot recall the information needed to compete it. Any advice is appreciated!
Thanks, Roman.

Answer 1

After flipping the coordinates of Barfield to I think fix the problem, I get

`d = 8-7/{tan 43^circ } approx 0.4934 .`

Explanation:

I spent a week in Barfield one night.

This problem seems a bit misstated. If Barfield was 7 km north, 0 km east of Westgate, that would require a bearing, usually meaning the angle relative to due north, of `0^circ`. As long as the bearing angle is less than `45^circ` we'd be going more north than east, so that's where Barfield should be, but it isn't. I'm going to assume we meant that Barfield is 8 km north and 7 km east of Westgate.

Let's start with a figure. I'll use the cartesian plane like a map, with up being north and right being east. I put Westgate at the origin `W(0,0)` and Barfield at `B(7,8)` and drew the segment. I wrote `41.2^circ` for the angle between the segment and the y-axis, complementary to the usual labeling.

Then I drew a point `S(7, y),` `y` being around `7.5,` drew the segment WS, and labeled the y axis angle `43^circ.`
.
According to the picture:

`tan 41.2^circ = 7/8`

We can check that with a calculator

`tan41.2^circ - 7/8 approx 0.000433823 quad` Close enough

It seems if we've understood bearings correctly our restatement was correct.

`tan 43^circ = 7/y`

`y = 7/tan 43^circ`

The distance we seek is

`d = 8-y = 8-7/{tan 43^circ } approx 0.4934 .`

That was a pretty good guess drawing `y` at `7.5.`