Question

question 1 consider the vector a = 52 m pointed 75 degrees measured ccw from the positive x axis. determine the x - component of vector a. add your answer integer, decimal, or e notation allowed question 2 consider the vector a = 52 m pointed 75 degrees measured ccw from the positive x axis. determine the y - component of vector a. add your answer integer, decimal, or e notation allowed

Explanation:

Step1: Recall vector - component formula

The x - component of a vector $\vec{A}$ with magnitude $A$ and direction $\theta$ (measured counter - clockwise from the positive x - axis) is given by $A_x=A\cos\theta$, and the y - component is given by $A_y = A\sin\theta$. Here, $A = 52$ m and $\theta=75^{\circ}$.

Step2: Calculate the x - component for Question 1

$A_x=A\cos\theta=52\times\cos(75^{\circ})$. Since $\cos(75^{\circ})=\cos(45^{\circ} + 30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259$. So $A_x = 52\times0.259 = 13.468\approx13.47$ m.

Step3: Calculate the y - component for Question 2

$A_y=A\sin\theta=52\times\sin(75^{\circ})$. Since $\sin(75^{\circ})=\sin(45^{\circ}+30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966$. So $A_y=52\times0.966 = 50.232\approx50.23$ m.

Answer:

Question 1: 13.47
Question 2: 50.23