Hello deltaxray Originally Posted by

**deltaxray** A lighthouse B is 8km from a point A and its bearing from A is 70 degrees E of N. A ship sails from A for 6km on a course due NE and then alters its course to 50 degrees E of N. Find from the tables its distances N and E of A when it is due North of B.

Im not sure how to do this questions. Thanks

I'm not sure what 'from the tables' means. But here's how I should do it.

B is $\displaystyle 8 \sin 70^o$ km East of A and $\displaystyle 8\cos 70^o$ km North of A. (Check this out on a diagram. I'll leave you to work these values out.)

When the ship has completed the first $\displaystyle 6$ km leg, it is $\displaystyle 6\sin45^o$ km East and $\displaystyle 6\cos45^o$ km North of A. (Work these out as well.)

If it then sails a distance $\displaystyle x$ km on a bearing $\displaystyle 050^o$, it will travel a further distance $\displaystyle x\sin50^o$ km East and $\displaystyle x \cos 50^o$ km North.

If the ship is now due North of B, its distance East of A is the same as the distance that B is East of A. So:

$\displaystyle 6\sin 45^o + x\sin50^o = 8\sin 70^o$

From this we can work out the value of $\displaystyle x$ and hence how far North it is of A. (You've already worked out how far East it is.)

Can you complete the working now?

Grandad