Question 1117405: A touring boat was heading toward an island 80 nautical miles due south of where it left port.after travelling 15 nautical miles,it headed 8 degree east of south to avoid a fleet of commercial fishermen.Atertravelling 6 nautical miles,it turned to head directly toward the island.How far was the boat from the island at the time it turned?
(i)can a diagram help you understand the problem?
(ii)what are you asked to find?
(iii)which measurements do you need to solve the problem?
The question is from the chapter of law of sines and law of cosines
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Part (i): A diagram isn't 100% needed though it helps visualize the problem.
If you are a visual learner, then I suggest you make a diagram.
For me, if I get stuck on a problem like this, a diagram usually helps things click a lot easier than words alone.
This is what the diagram should look like
The points are as follows:
Point A = port location (starting point for the boat)
Point B = point in which the boat turns 8 degrees east of south (see note1 below)
Point C = second turning point after which the boat travels directly to the island
Point D = the island location (ending point)
The boat travels from A, to B, to C, then to D in that order.
Given/Known Distances:
AD = 80 nautical miles
AB = 15 nautical miles
BC = 6 nautical miles
BD = 65 nautical miles (see note2)
note1: saying "8 degrees east of south" means you start facing directly south and then turn 8 degrees toward east as the diagram shows.
You may see "8 degrees east of south" written as  which is shorthand notation saying "face south, turn 8 degrees toward east".
note2: We aren't given the distance from B to D, but we can use the fact that AB+BD = AD which becomes BD = AD-AB. So, BD = 80 - 15 = 65. Or you can note how 15+65 = 80.
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Part (ii)
We're asked to find the distance from point C to point D. This is the distance after the boat makes a second turn and travels directly to the island without any more turning. For now that distance is unknown so we'll label it x.
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Part (iii)
We must use the law of cosines to find the value of x. It's not possible to use the law of sines as there isn't enough info. We would need to know either angle D or angle C to be able to determine x using the law of sines.
The measurements we'll use are
BD = 65 miles
BC = 6 miles
angle B = 8 degrees (focus solely on triangle BCD)
Side b is opposite angle b, so b = x
Let a = 6 and c = 65 be the two known sides of the triangle
So we have
a = 6
b = x
c = 65
angle B = 8 degrees
Plug the values into Law of Cosines formula and solve for x. Make sure your calculator is in degree mode.
b^2 = a^2 + c^2 - 2*a*c*cos(B)
x^2 = 6^2 + 65^2 - 2*6*65*cos(8)
x^2 = 36 + 4225 - 780*cos(8)
x^2 = 36 + 4225 - 780*0.99026806874157
x^2 = 36 + 4225 - 772.409093618425
x^2 = 3488.59090638158
x = sqrt(3488.59090638158)
x = 59.0642946828418
The distance from point C to point D is roughly 59 nautical miles if you round to the nearest whole number.
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