Question

vector $vec{b}$ has magnitude 1.00 and makes an angle of 70 degrees above the positive x-axis. vector $vec{d}$ has magnitude 3.40 and makes an angle of 22 degrees above the negative x-axis. what is the magnitude of their resultant $vec{b} + vec{d}$?

Explanation:

Step1: Find $\vec{B}$ components

$B_x = 1.00\cos(70^\circ) \approx 0.342$, $B_y = 1.00\sin(70^\circ) \approx 0.940$

Step2: Find $\vec{D}$ components

$D_x = 3.40\cos(158^\circ) \approx -3.17$, $D_y = 3.40\sin(158^\circ) \approx 1.27$

Step3: Sum x-components

$R_x = B_x + D_x = 0.342 - 3.17 \approx -2.83$

Step4: Sum y-components

$R_y = B_y + D_y = 0.940 + 1.27 \approx 2.21$

Step5: Calculate resultant magnitude

$|\vec{B}+\vec{D}| = \sqrt{R_x^2 + R_y^2}$

Answer:

$\sqrt{(-2.83)^2 + (2.21)^2} \approx 3.60$