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?

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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!

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Answer 1 · 2018-04-02T22:38:33

BC = 18km
Beating of C from B is 172.8

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