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 $ˆ A$ $hatA$ C is 147 -36 = 111

Using the cosine rule : $a^{2} = b^{2} + c^{2} - 2 b c (cos A)$ $a^2 = b^2 + c^2 -2bc(cosA)$

$B C^{2} = {8.4}^{2} + {13.2}^{2} - 2 \times 8.4 \times 13.2 \times cos (111)$ $BC^2 = 8.4^2 +13.2^2 -2 xx 8.4 xx 13.2 xx cos(111)$

$B C^{2} = 324.2716765$ $BC^2 = 324.2716765$

#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 $ˆ B$ $hatB$ C from 216

Cos (B) = $\frac{18^{2} + {8.4}^{2} - {13.2}^{2}}{2 \times 8.4 \times 18}$ $[18^2 + 8.4^2 - 13.2^2]/(2 xx 8.4 xx 18)$

Cos B = $\frac{51}{70}$ $51/70$

B = ${cos}^{- 1}$ $cos^(-1)$ $(\frac{51}{70})$ $(51/70)$

B= 43.23323481

B= 43.2

So the bearing is 216 - 43.2 = 172.8