Question

Vector A is 21 units long and is at an angle of 84° counterclockwise from east. Vector B is a length of 8 units and an angle of -130⁰ clockwise. What is the direction of -4.3A+7.2B to the nearest tenth of degree?​

Answer 1

-78.4°

Step-by-step explanation:

`A_(angle) = 180 + 84 = 264\\A = < A_xcos(264),A_ysin(264) > = < 21cos(264),21sin(264) > \\B_(angle) = 130\\B= < B_xcos(130),B_ysin(130) > = < 8cos(130),8sin(130) > \\-4.3A+7.2B= < -4.3*21*cos(264)+7.2*8*cos(130),-4.3*21*sin(264)+7.2*8*sin(130)\\-4.3A+7.2B= < -27.586,133.929 > \\Angle = tan^(-1)(133.929/-27.586) = -78.4`