Question 175045
I'll convert the velocity vectors to distance vectors. Then 
I'll find the points on the x-y plane where the end up and use the distance formula
" One travels in the direction 60 degrees east of north."
That's the same as 30 degrees N of E. In a half-hour this car
would have gone 60 km/hr x 1/2 hr = 30 km
The component along the y-direction is
{{{15}}}
The component along the x-direction is 
{{{15/sqrt(3)}}}
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" The other travels in the direction 45 degrees east of south."
That's the same as 45 degrees south of east
This car also would have gone 60 km/hr x 1/2 hr = 30 km
The component along the y-direction is
{{{-30/sqrt(2)}}}
The component along the x-direction is 
{{{30/sqrt(2)}}}
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Now I have 2 points
({{{15}}},{{{15/sqrt(3)}}})
and
({{{30/sqrt(2)}}},{{{-30/sqrt(2)}}})
Now I use the distance formula
{{{d = sqrt((30/sqrt(2) - 15)^2 + (-30/sqrt(2) - 15/sqrt(3))^2)}}
{{{d = sqrt((21.21 - 15)^2 + (-21.21 - 8.66)^2)}}}
{{{d = sqrt(6.21^2 + 29.87^2)}}}
{{{d = sqrt(38.56 + 892.22)}}}
{{{d = sqrt(930.78)}}}
{{{d = 30.51}}} km