Bearings: A radar station picks up a distress call from a ship. It is 50km away on a

[adi9891]

Hi guys,

So I'm having a bit of trouble answering this particular question that involves bearings.

A radar station picks up a distress call from a ship. It is 50km away on a bearing of 341°. The station contacts a lifeboat at sea which is 20km away on a bearing of 213°. Draw a diagram and using it find the distance and bearing of the ship from the lifeboat.

I have drawn this out, but somehow I can't seem to figure out how to find the distance or the bearing. Somehow I think the bearing from the lifeboat to the ship is 341 but I have no idea how to prove it. As for the distance, I can't seem to figure out how to determine it.

Any help would be really appreciated

Thanks,
Adi

 

[Johulus]

Hi guys,

So I'm having a bit of trouble answering this particular question that involves bearings.

A radar station picks up a distress call from a ship. It is 50km away on a bearing of 341°. The station contacts a lifeboat at sea which is 20km away on a bearing of 213°. Draw a diagram and using it find the distance and bearing of the ship from the lifeboat.

I have drawn this out, but somehow I can't seem to figure out how to find the distance or the bearing. Somehow I think the bearing from the lifeboat to the ship is 341 but I have no idea how to prove it. As for the distance, I can't seem to figure out how to determine it.

Any help would be really appreciated

Thanks,
Adi

So, as you can see imaginary lines that connect lifeboat with the station and ship with the station and line that represents the distance between the ship and the lifeboat form a triangle. Let's say: \(\displaystyle a=50 km \qquad b=20km \qquad c...distance \qquad B...relative \, \, bearing \) . If you subtract the absolute bearing of the lifeboat from the absolute bearing of the ship you get \(\displaystyle \gamma \): \(\displaystyle \gamma=341°-213°=128°\). Now you can get the distance between the ship and the lifeboat using cosine theorem: \(\displaystyle c^2=a^2+b^2-2ab\cos\gamma \).
Now you should find \(\displaystyle \alpha \) in order to find the bearing of the ship from the lifeboat. In order to do that you can apply sine theorem: \(\displaystyle \dfrac{a}{\sin\alpha}=\dfrac{c}{\sin\gamma} \). Once you have \(\displaystyle \alpha \), it is easy to find B(relative bearing).

And absolute bearing of the ship from the lifeboat A should be: \(\displaystyle A=B \, (relative \, \, bearing) \, + 213° \) .

 

[adi9891]

Thanks so Johulus for taking the time to do that for me. Very kind of you

Now to go ahead and take the credit as if I figured this out...