this question I have no idea how to do so if you could help me please.
The second narrows in vancouver, British Columbia, is 3.4 km from Deep Cove on a bearing of 49 degrees. A boat leaves second narrows on a bearing of 261 degrees. after sailing for 1.3 km, the boat turns and travels 2.4 km to reach Deep Cove. to the nearest degree, in what direction is the boat now sailing?
a)232 degrees b)247 degrees c)103 degrees d)212 degrees
e) 218 degrees
2007-01-16 15:18:50 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
There's too much information here, I hope it's all consistent.
Because three distances are given (DC to SN 3.4, SN to turn point 1.3, and TP to DC 2.4) we can use the cosine rule to find the angle between the two sides at SN:
Cos S = (3.4^2 + 1.3^2 -2.4^2)/(2*3.4*1.3)
S = 32.08 degrees which is correct, as checking out the bearings makes it 32 degrees exactly.
Use the sine rule to find the angle between the two sides at DC:
sin D = 1.3*sin (32 deg)/2.4
I get 16.68 degrees for that, and subtracting it from 49 gives
32.32 degrees.
Since the boat is sailing back towards that point, its bearing is
180 + 32.32 = 212.32 degrees, so answer d is the one.
PS Have you a diagram for this? I've just done one in Paint, but when I tried to send it, I got a message that I first have to enable Yahoo Answers email, and I don't know how to do that. If you email h_chalker@yahoo.com.au I'll send it to you.
2007-01-16 15:58:31 · answer #1 · answered by Hy 7 · 0⤊ 0⤋
The answer is d) 212°
The hardest part of this is to coordinate bearings and angles. I used the Law of Sines.
This is difficult to explain without a diagram.
2007-01-17 01:38:02 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
not sure 100% but i think A
2007-01-16 23:43:33 · answer #3 · answered by no way jose 1 · 0⤊ 0⤋