Question 1044501
|=========Pt 1 (approx)
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Start=================Pt 2 (approx)


It goes NE (45 deg bearing) for 5 km.
x value is 5 cos 45 or (5/2) sqrt (2)
y value is 5 sin 45 or (5/2) sqrt (2)
the other vector bears 120 degrees
x value is 7 cos (-30)=(7/2)sqrt (3)
y-value is 7 sin (-30)=-3.5
add the two
x value is (5/2) sqrt (2)+(7/2) sqrt (3) or 9.598
y value is (5/2) sqrt (2)-3.5 or 0.036
The distance is the sqrt of each of those squared.
Numerically it is 9.60 km (rounding only at end)
Direction is tan (-1) (y/x)=0.036/9.598=0.214 deg above the x-axis or 90.2 deg true
9.6 km at 90 degrees.