Question
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)
1. Calculate the nearest km, the distance between Belleville (B) and comptin (C)
Answers
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
= ....
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
= ....
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.
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.
Kimoya
Part two of that same questions said : calculate to the nearest degree the measure of ABC
henry2,
2. sinB/310 = sin60/276.
sinB = 310*sin60/276,
B =
sinB = 310*sin60/276,
B =
Lyra
294