Question

Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30°. What is the direction angle of their vector sum?

Answer 1

Answer: B. 2.20°

Step-by-step explanation: I got this correct on Edmentum.

Answer 2

keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.


`\bf u= \begin{cases} x=7cos(330^o)\\ \qquad 7\cdot (√(3))/(2)\\ \qquad (7√(3))/(2)\\ y=7sin(330^o)\\ \qquad 7\cdot -(1)/(2)\\ \qquad -(7)/(2) \end{cases}\qquad \qquad v= \begin{cases} x=8cos(30^o)\\ \qquad 8\cdot (√(3))/(2)\\ \qquad (8√(3))/(2)\\ y=8sin(30^o)\\ \qquad 8\cdot (1)/(2)\\ \qquad 4 \end{cases}`


`\bf u+v\implies \left( (7√(3))/(2),-(7)/(2) \right)+\left( (8√(3))/(2),4 \right)\implies \left( (7√(3))/(2)+(8√(3))/(2)~~,~~ -(7)/(2)+4\right) \\\\\\ \left(\stackrel{a}{(15√(3))/(2)}~~,~~ \stackrel{b}{(1)/(2)}\right)\\\\ -------------------------------`


`\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{(1)/(2)}{(15√(3))/(2)}\implies tan(\theta )=\cfrac{1}{15√(3)} \\\\\\ \measuredangle \theta =tan^(-1)\left( \cfrac{1}{15√(3)} \right)\implies \measuredangle \theta \approx 2.20422750397203^o`