Two particles are moving in a plane; one has the velocity component vix=1m/s, viy=5m/s and the other has the velocity components v2x=4m/s and v2y=3m/s if both started from the same point what is the angle between their paths

Answer 1

calculate the sines and cosines of the angles ....

sintheta1=5/sqrt26
cosTheta1=1/sqrt26

sinTheta2=3/5
cosTheta2=4/5 check all those...

Now,
Sin(A-B)=sianAcosB - CosBsinA

then compute aine (A B), and finally, the arcsin(A-B) which is the angle

Answer 2

V1 = 1 + 5i = 5.1m/s[78.7o].

V2 = 4 + 3i = 5m/s[36.9o].

78.7 - 36.9 = 41.8o between V1 and V2.

Answer 3

Both solutions of the impersonator of MathMate and Henry give the same correct answer of 41.820°.

By the way, for your information, impersonation is a serious offence at Jiskha, liable to be banned.