a+b+c = (42cos27,42sin27) + (42cos194, 42sin194) + (42cos312,42sin312)
= (24.7733, -22.3052)
magnitude = √(24.7733^2 + 22.3052^2) = appr 33.335
direction angle Ø, such that
tanØ = -22.3052/24.7733
I get Ø = appr -41.998°
or 318.00°
compare with your work, and find your error.
Three vectors a, b, and c, each have a magnitude of 42 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 27 ˚, 194 ˚, and 312 ˚, respectively. B) i need the angle of the vector a+b+c (relative to the +x direction in the range of (-180°, 180°) i got 42 but its not correct do i have to add it to 180? D)angle of a-b+c in the range of (-180°, 180°) i got 5 degrees for this. and it is incorrect
5 answers
how is -41.998 = to 318?
Displacement = 42m[27] + 42[194] + 42[312].
X = 42*Cos27 + 42*Cos194 + 42*Cos312 = 24.77 m.
Y = 42*sin27 + 42*sin194 + 42*sin312 = -22.3 m.
X + Yi = 24.77 - 22.3i = 33.33m[-42o] = 33.33m[42o] S. of E. = 33.33m[318o]CCW.
Angle calculations:
Tan A = (-22.30/24.77 = -0.90028,
A = -42o, Q4.
360-42 = 318o CCW from +x-axis.
X = 42*Cos27 + 42*Cos194 + 42*Cos312 = 24.77 m.
Y = 42*sin27 + 42*sin194 + 42*sin312 = -22.3 m.
X + Yi = 24.77 - 22.3i = 33.33m[-42o] = 33.33m[42o] S. of E. = 33.33m[318o]CCW.
Angle calculations:
Tan A = (-22.30/24.77 = -0.90028,
A = -42o, Q4.
360-42 = 318o CCW from +x-axis.
so, if the x-component is positive and the y-component is negative it will end up in the fourth quadrant? therefore, I would always have to subtract it by 360?
nvm i got it thanks
1 answer