Question

B is on a bearing 36 degrees from A and 8.4km from A. C is on a bearing of 147 degrees from A and 13.2km from A. How to find: (i) BC? (ii) the bearing of C from B?

I was able to do part (i) and got 18km. The bearings part is what I don't really understand. I can work out 'easier' bearings questions, but this one has me stuck.
Thanks!

Answer 1

BC = 18km
Beating of C from B is 172.8

Explanation:

If C is on a bearing of 147 and B is on a bearing 36 then B`hatA`C is 147 -36 = 111

Using the cosine rule : `a^2 = b^2 + c^2 -2bc(cosA)`

`BC^2 = 8.4^2 +13.2^2 -2 xx 8.4 xx 13.2 xx cos(111)`

`BC^2 = 324.2716765`

#BC = 18.0075449

The bearing of A from B is 216. To work out the bearing of C from B we have to subtract angle A`hatB`C from 216

Cos (B) = `[18^2 + 8.4^2 - 13.2^2]/(2 xx 8.4 xx 18)`

Cos B = `51/70`

B = `cos^(-1)` `(51/70)`

B= 43.23323481

B= 43.2

So the bearing is 216 - 43.2 = 172.8