Okay, let's do math.
For the total x displacement:
For the first part of the man's walk:
He walked 32.5 km in a 45° direction.
To find the x-displacement, cos θ = adjacent side/hypotenuse.
So after some variable manipulation, adjacent side = cos θ * hypotenuse
θ = 45°
hypotenuse = 32.5 km
Plug them in, and you get 22.98 km
For the second part of his walk, he travel 15.5 km east. Since there is no change in y when traveling solely east, change in x for this instance is 15.5 km
22.98 km + 15.5 km = total x-displacement, which is 38.48 km.
Now to solve for the y-displacement:
sin θ = opposite side/hypotenuse.
(the opposite side is the y-displacement)
After some variable manipulation, opposite side (y-displacement) = sin θ * hypotenuse
θ = 45°
hypotenuse = 32.5 km
Plug them in, and you should get 22.98 km for y-displacement in that instance.
In the second instance, the man travels solely east. Therefore, there is no y-displacement, so y = 0
22.98 km + 0 km = 22.98 km total y-displacement!
Now, after getting both total displacements, you can solve both of your questions.
A^2 + B^2 = C^2
A=X, Y=B, and C=hypotenuse
So, X^2 + Y^2 = h^2
After some variable manipulation, h = √(X^2 + Y^2)
X = 38.48 km
Y= 22.98
Plug them in, and you should get 44.82 km for your magnitude.
You also need to find direction, right? You already have the total X and Y displacement, so that's easy.
tan θ = opposite side (y)/adjacent side(x)
After some variable manipulation, θ = tan^ -1 (opposite side (y)/ adjacent side (x) )
Plug your total displacements in, and θ = 30.84° north of east.