The Student Room Group

cosine rule again!

A rally car leaves its base on a bearing of 030degrees and drives in a desert for 12km. It then turns and drives a further 8km on a bearing of 100degrees after which it breaks down.

a) Draw a clear sketch of this journey, stating all known lengths and angles.
b) A recovery vehicle leaves base to pick up the car. How far must it travel and on what bearing should it head?


Im having problems with the diagram , i hope someone can help me with the diagram because then part b should be straight forward
Reply 1
dec0!
A rally car leaves its base on a bearing of 030degrees and drives in a desert for 12km. It then turns and drives a further 8km on a bearing of 100degrees after which it breaks down.

a) Draw a clear sketch of this journey, stating all known lengths and angles.
b) A recovery vehicle leaves base to pick up the car. How far must it travel and on what bearing should it head?


Im having problems with the diagram , i hope someone can help me with the diagram because then part b should be straight forward

see diagram attached
Reply 2
it seems that u have problems with the "bearing"! dont u?
Reply 3
ok, i got that far as well, but how do u work out the inner angles?:s-smilie:
Reply 4
I'm not sure but is it something like this:
isthisright.JPG
Reply 5
well see attachment...


other angles could be found using sine rule...
Reply 6
ESBay
I'm not sure but is it something like this:
isthisright.JPG


i liked ur way of thinking! it is correct...
Reply 7
yazan_l
i liked ur way of thinking! it is correct...

Thanks for that :smile:
Reply 8
ok thanks a lot for all ur help dudez! i get it now,nice clearly explained ESBay, but u still got the answer wrong u must hav gone somewhere wrong in ur calculation, i got 16.5, the same as the back of the book :biggrin:
Reply 9
x^2 = b^2 + c^2 - 2bc cosX
x^2 = 12^2 + 8^2 - 2x12x8xcos110
x^2 = 208-192cos110
x^2 = 273.66......
x = 16.5 km