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.

1 Answer
Apr 26, 2018

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.#