A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of south.
I know the formulas but having trouble figuring out the degree of angle.
Ax=4(cos30)= 3.4641
Ay=4(sin30)= 2
Bx=3.7(cos20)= 3.477
By=3.7 (sin20)= 1.2654
Rx = 3.4641 + 3.477 = 6.9411
Ry = 2+1.2654 = 3.2654
R=7.6708
tan theta = Rx/Ry
= 6.9411/3.2654
=2.125650762540577
=0.037116634989 SW
Please let me know if I am doing this correct
R=7.2
49 degrees south of west
I know this is dead, but to find the degree of an angle after you solved for Tan(theta), you just do Tan^-1(2.125650762540577) which would be 64.80563. That is the degree. Hope this helps that one random person who sees this one day.
Thanks Jacob's awesome!
You are on the right track in using the correct formulas to find the x and y components of the vectors. However, there are some mistakes in the calculations.
To find the x and y components of the first vector:
Ax = 4.0 * cos(30°) = 3.4641 m/s (round to four decimal places)
Ay = 4.0 * sin(30°) = 2.0 m/s
To find the x and y components of the second vector:
Bx = 3.7 * cos(20°) = 3.4666 m/s (round to four decimal places)
By = 3.7 * sin(20°) = 1.2679 m/s (round to four decimal places)
Now, adding the x and y components:
Rx = 3.4641 + 3.4666 = 6.9307 m/s (round to four decimal places)
Ry = 2.0 + 1.2679 = 3.2679 m/s (round to four decimal places)
To find the magnitude of the resultant vector R:
R = sqrt(Rx^2 + Ry^2)
= sqrt((6.9307)^2 + (3.2679)^2)
= sqrt(47.99744 + 10.67268)
= sqrt(58.67012)
= 7.6667 m/s (round to four decimal places)
Now, to find the angle of the resultant vector:
tanθ = Ry / Rx
θ = tan^(-1)(Ry / Rx)
= tan^(-1)(3.2679 / 6.9307)
= tan^(-1)(0.4710)
= 25.98° (rounded to two decimal places)
Therefore, the angle of the resultant vector is approximately 26°.