We have triangle with two known sides, a =27.1 km and b =75 km, and the angle α between them
α = 90 º - 16 º -26 º = 48º.
Let’s use the law of cosines to solve for the third side (magnitude of displacement)
r = sqrt(a² +b²-2•a•b•cosα) = sqrt(75² +27.1² - 2•75•27.1•cos48º) = 60.34 km.
Then use the law of sines to find the angles opposite to side a = 27.1 km.
a/sinθ = r/sinα .
sin θ = sinα •a/r =sin48•27.1/60.34 = 0.333.
θ = 19.6º,
φ = 90 º – (26 º -19.5 º) = 90 º – 6.5 º = 83.5 º (N of E)
v(average) = r/t = 60.34•60/52 = 69.62 km/hr
Its direction will be the same as the direction of displacement.