Question

4) From A, B lies 11 km away on a bearing of 0410 and C lies 8 km away on a bearing of 3410. Find

a) The distance between B and C.

b) The bearing of B from C.

Answer 1

From the figure you can see that the angle between C and B (green angle) is equal to

(360$\degree$ -341$\degree$ ) + 41$\degree$ = 60$\degree$

We get the triangle, let call side AB - c, AC-b, BC-a

1)

So,using the law of cosines:

$a^2 = b^2 + c^2 - 2 bc\times cos A\\ a^2 = 8^2 + 11^2 - 2 (8)(11)\times cos 60\degree\\ a^2 = 97,\\ a = 9.8 km$

BC=9.8 km

2)

From the law of Sines find the angle C at point C

$\frac{9.8}{sin60 \degree} = \frac{11}{sin C}\\ sin C = \frac{sin60 \degree \times 11}{9.8}=0.972\\ C=sin^{-1}(0.972 )= 75.3 \degree$

The bearing of B from C is the angle formed by the line joining C and B and rotating about C. By Geometry this angle is

180° - (C + 19°) = 180°-(75.3°+19°)=85.7 $\degree$