Trignometry!! (1 Viewer)

[hobozforlife]

A yacht sails from Port A to Port B 18 nautical miles on a bearing 150 degrees T. At port B it changes direction and heads to port C on a bearing of 247 degrees T at a distance of 25 nautical miles.

- Find the size of angle ABC and hence find the distance of port C to port A.
- Find the compass bearing and true bearing from port C to port A.

 

[gr_111]

Diagrams are very helpful in these sorts of questions...

i)After finding angle ABC as shown in diagram, use the cosine rule to determine distance CA, or x.



ii) The acute angle between the grey line and the black line at C must equal 67°, because it is co-interior to the 83° and 30° at B.
We can then use the sine rule to divide the 67° into the angle in the triangle, and therefore work out the bearing angle.

<a href="http://www.codecogs.com/eqnedit.php?latex=\text{Let } y = \angle ACB\\ \\ \frac{siny}{18} = \frac{sin83}{29}\\ \\ siny = \frac{18sin83}{29}\\ \\ y = 38.03....\\ \therefore \angle ACB \approx 38 \text{ degrees}\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\text{Let } y = \angle ACB\\ \\ \frac{siny}{18} = \frac{sin83}{29}\\ \\ siny = \frac{18sin83}{29}\\ \\ y = 38.03....\\ \therefore \angle ACB \approx 38 \text{ degrees}\\" title="\text{Let } y = \angle ACB\\ \\ \frac{siny}{18} = \frac{sin83}{29}\\ \\ siny = \frac{18sin83}{29}\\ \\ y = 38.03....\\ \therefore \angle ACB \approx 38 \text{ degrees}\\" /></a>

The true bearing of A from C is then and the compass bearing is something like (if i recall compass bearings correctly!)