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
= ....
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)
5 answers
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.
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.
Part two of that same questions said : calculate to the nearest degree the measure of ABC
2. sinB/310 = sin60/276.
sinB = 310*sin60/276,
B =
sinB = 310*sin60/276,
B =
294
2 answers