Acceleration/Velocity vectors question

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An object experiences an acceleration of 2.3i + 3.6j m/s² over a 10-second interval, resulting in a final velocity of 33i + 15j m/s. The initial velocity was calculated as 10i - 21j m/s. The change in speed was determined by subtracting the magnitudes of the initial and final velocity vectors, yielding a result of 13 m/s. The change in direction was found using vector graphs and the law of sines, resulting in a total direction change of 88.9º. The calculations for all parts of the homework question were confirmed to be correct.
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ok, homework question
An object undergoes acceleration of 2.3i+3.6j m/s^2 over a 10s interval. At the end of this time, its velocity is 33i+15j m/s.
A. What was its velocity at the beginning of the 10s interval?
B. By how much did its speed change?
C. By how much did its direction change?

I know how to do part A, as I did it like this:
(2.3i+3.6j)10=23i+36j ; 33i+15j-23i-36j=10i-21j
I am stumped on parts 2 and 3, I need help on where to begin on those.
 
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ok, now i just found part c. I just set up the vector graphs, and did the law of sines to find the angles at the origin. They came out to be 24.4 and 64.5, so added together to get the total direction change, I came out with 88.9º. In the back of the book, that is correct. Now, i should be able to get part b, so we'll see.
 
ok, now i found part b. I finally realized that all I had to do was take the magnitudes of the velocity vectors and subtract them. Came out to 13 m/s, which is the correct answer. I guess I just needed to take the time and graph em, haha...dont need the help...yet.
 
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