Asked by Kimoya

A ship travels from Akron A on a bearing of 030° to Belleville (B) 90km away. It then travels to comptin (C) which is 310km due east of Akron (A)
1. Calculate the nearest km, the distance between Belleville (B) and comptin (C)
Answered by Reiny
Did you make a sketch?

Looks to me that you have a triangle ABC, with AB=30, AC=310 and angle A = 60 degrees
A clear case of the cosine law:
BC^2 = 30^2 + 310^2 - 2(30)(310)cos 60
= ....
Answered by henry2,
All angles are measured CW from +y-axis.
Given: AB = 90km[30o], BA = 90km[30+180] = 90km[210o].
AC = 310km[90o].

BC = BA + AC = 90[210o] + 310[90o],
BC = (90*sin210+310*sin90) + (90*cos210+310*cos90)i,
BC = 265 - 78i = 276km[-74o] = 276km[106o]CW.
Answered by Kimoya
Part two of that same questions said : calculate to the nearest degree the measure of ABC
Answered by henry2,
2. sinB/310 = sin60/276.
sinB = 310*sin60/276,
B =
Answered by Lyra
294
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