Asked by Anonymous
Vector Barrowbold has magnitude 9.4 and direction 14° below the +x-axis. Vector Carrowbold has x-component Cx = -1.6 and y-component Cy = -7.7.
Compute direction of Carrowbold & direction of Carrowbold + Barrowbold.
Also direction of Carrowbold − Barrowbold and direction in ° counter-clockwise from the +x-axis!
Thnk you
Compute direction of Carrowbold & direction of Carrowbold + Barrowbold.
Also direction of Carrowbold − Barrowbold and direction in ° counter-clockwise from the +x-axis!
Thnk you
Answers
Answered by
Henry
C = -1.6 - i7.7. Q3.
tanAr = Y/X = -7.7 / -1.6 = 4.8125,
Ar = 78.3 Deg. =Reference angle.
A = 180 + Ar = 180 + 78.3 = 258.3 Deg.
B = 9.4 @ (-14) Deg. C = -1.6 -i7.7.
X = 9.4*cos(-)14 + (-1.6) = 7.52.
Y = 9.4*sin(-14) + (-7.7) = -9.97.
tanAr = Y/x = -9.97 / 7.52 = -1.3263,
Ar = -53 Deg.
A = -53 + 360 = 307 Deg.
C = -1.6 -i7.7. B = 9.4 @(-14) Deg.
X = -1.6 - 9.4*cos(-14) = -10.72.
Y = -7.7 - 9.4*sin(-14) = -5.43.
tanAr = Y/X = -5.43 / -10.72 = 0.5062.
Ar = 26.8 Deg.
A = 180 + 26.8 = 206.8 Deg.
tanAr = Y/X = -7.7 / -1.6 = 4.8125,
Ar = 78.3 Deg. =Reference angle.
A = 180 + Ar = 180 + 78.3 = 258.3 Deg.
B = 9.4 @ (-14) Deg. C = -1.6 -i7.7.
X = 9.4*cos(-)14 + (-1.6) = 7.52.
Y = 9.4*sin(-14) + (-7.7) = -9.97.
tanAr = Y/x = -9.97 / 7.52 = -1.3263,
Ar = -53 Deg.
A = -53 + 360 = 307 Deg.
C = -1.6 -i7.7. B = 9.4 @(-14) Deg.
X = -1.6 - 9.4*cos(-14) = -10.72.
Y = -7.7 - 9.4*sin(-14) = -5.43.
tanAr = Y/X = -5.43 / -10.72 = 0.5062.
Ar = 26.8 Deg.
A = 180 + 26.8 = 206.8 Deg.
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