A ship at A is to sail to C, 56km north and 258km east of A. After sailing N25°10’E for 120mi to P, the ship is headed toward C. Find the distance of P from C and the required course to mean C.
2 answers
Why is C considered "mean"?
The main part of the question is to make a decent diagram.
We can find AC using Pythagoras and I found it to be
264.00 km
Also using right-angled trig, AC makes an angle of
12.246° which makes angle PAC = 52.587°
In triangle PAC , we have AP = 120 , AC = 264
adn angle PAC = 52.587
by the cosine law:
PC^2 = 120^2 + 256^2 - 2(56)(256)cos52.587°
I will let you handle the drugery.
Once you have PC, you can use the Sine Law to find the angle at P.
From there it should be easy to find the bearing needed to get to C
We can find AC using Pythagoras and I found it to be
264.00 km
Also using right-angled trig, AC makes an angle of
12.246° which makes angle PAC = 52.587°
In triangle PAC , we have AP = 120 , AC = 264
adn angle PAC = 52.587
by the cosine law:
PC^2 = 120^2 + 256^2 - 2(56)(256)cos52.587°
I will let you handle the drugery.
Once you have PC, you can use the Sine Law to find the angle at P.
From there it should be easy to find the bearing needed to get to C
1 answer