Two people leave from the same spot and walk at 4 ft/sec going north and 5 ft/sec northwest. At 30 seconds, how fast is the distance between them changing??

at a time of t sec after they started walking, their distances are 4t and 5t ft, and the angle between them is 48 degrees

by the cosine law:
x^2 = 16t^2 + 25t^2 - 2(4t)(5t)sqrt(2)/2
x^2 = 41t^2 - 20sqrt(2)t^2
2x dx/dt = 82t - 40sqrt(2)t
when t= 30
x^2 = 41(900) - 20sqrt(2)(900)
x = 106.977...
dx/dt = (82(30) - 40sqrt(2)(30))/(213.95...)
= 3.566 ft/s

check my arithmetic