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All angles measured from the positive x-axis \cdot Vector1 \textendash north of east \cdot Vector2 \textendash units due east $\vec{C} = \vec{A} + \vec{B}$ $\theta_A = \tan^{-1}\frac{A_y}{A_x} = ____$ $\theta_B = \tan^{-1}\frac{B_y}{B_x} = ____$ $C_x = A_x + B_x = ____$ $C_y = A_y + B_y = ____$ $|C| = \sqrt{C_x^2 + C_y^2} = ____$ $\theta_C = \tan^{-1}\frac{C_y}{C_x} = ____$ All angles measured from the positive x-axis \cdot Vector1 \textendash north of east \cdot Vector2 \textendash units due east $\vec{C} = \vec{A} + \vec{B}$ $\theta_A = \tan^{-1}\frac{A_y}{A_x} = ____$ $\theta_B = \tan^{-1}\frac{B_y}{B_x} = ____$ $C_x = A_x + B_x = ____$ $C_y = A_y + B_y = ____$ $|C| = \sqrt{C_x^2 + C_y^2} = ____$ $\theta_C = \tan^{-1}\frac{C_y}{C_x} = ____$

          All angles measured from the positive x-axis
\cdot Vector1 \textendash north of east
\cdot Vector2 \textendash units due east
$\vec{C} = \vec{A} + \vec{B}$
$\theta_A = \tan^{-1}\frac{A_y}{A_x} = ____$
$\theta_B = \tan^{-1}\frac{B_y}{B_x} = ____$
$C_x = A_x + B_x = ____$
$C_y = A_y + B_y = ____$
$|C| = \sqrt{C_x^2 + C_y^2} = ____$
$\theta_C = \tan^{-1}\frac{C_y}{C_x} = ____$
All angles measured from the positive x-axis
\cdot Vector1 \textendash north of east
\cdot Vector2 \textendash units due east
$\vec{C} = \vec{A} + \vec{B}$
$\theta_A = \tan^{-1}\frac{A_y}{A_x} = ____$
$\theta_B = \tan^{-1}\frac{B_y}{B_x} = ____$
$C_x = A_x + B_x = ____$
$C_y = A_y + B_y = ____$
$|C| = \sqrt{C_x^2 + C_y^2} = ____$
$\theta_C = \tan^{-1}\frac{C_y}{C_x} = ____$

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All angles measured from the positive x-axis
·Vector1 –north of east
·Vector2 –units due east
C⃗ = A⃗ + B⃗
= tan^-1(Ay)/(Ax) =
= tan^-1(By)/(Bx) =
Cx = Ax + Bx =
Cy = Ay + By =
|C| = √(Cx^2 + Cy^2) =
= tan^-1(Cy)/(Cx) =
All angles measured from the positive x-axis
·Vector1 –north of east
·Vector2 –units due east
C⃗ = A⃗ + B⃗
= tan^-1(Ay)/(Ax) =
= tan^-1(By)/(Bx) =
Cx = Ax + Bx =
Cy = Ay + By =
|C| = √(Cx^2 + Cy^2) =
= tan^-1(Cy)/(Cx) =

Added by Sandra G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

09/09/2022

Step 1

- The texts provide various equations and measurements related to vectors and angles. - Some of the given equations include: - C = A + B - OA = tan-1(Ay/Ax) - Bx = Cx = A + B - Cy = Ay + By - c = tan-1(y/Cx) - 1C1 = c + c - c = tan-1(Cy/Cr) Show more…

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Texts: All angles measured from the positive x-axis. Vector1 - north of east. Vector2 - units due east. C = A + B. OA = tan-1(Ay/Ax). 0.01 Width. Height: 1.0 Width: 0.03 X. Bx = Cx = A + B. Height: 1.4 Width: 0.03 X. Cy = Ay + By. 1C1 = c + c. c = tan-1(y/Cx). All angles measured from the positive x-axis. Vector1 - north of east. Vector2 - units due east. Height: 0.52 Width: 0.92 X. Height: 0.69 Width: 0.03 X. Height: 1.46 Width: 0.63 X. C = A + B. O = tan-1(Ay/Ax). Bx = Cx = A + B. Cy = Ay + By. 1c = c + c. c = tan-1(Cy/Cr).

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00:02 In our problem the magnitude of vector a is given to be sixty six point two and its direction theta a is given to be twenty nine point three degrees with respect to the positive x axis and the magnitude of vector b is given to be forty one point four and its direction theta b is given to be fifty six point two degrees with respect to the negative x axis and the magnitude of vector c is given to be forty eight point nine and it is along the negative y axis using these values we can find both x x and y components of the three vectors.

00:59 Ax, the x component of a vector, can be written as a, cos, theta a.

01:06 Substituting the values, we get 66 .2, multiplied by cos, 29 .3 degrees.

01:14 On calculating, we get 57 .730, and its y component, a .y can be written as a, sine theta a, which is equal to 66 .2 times sine 29 .3 degrees.

01:33 On calculating, we get 32 .397.

01:39 And the x component of b vector can be written as minus b, cos theta b.

01:47 Here, the negative sign is due to the fact that theta is measured from the negative x axis.

01:57 Now on substituting the values we get minus of 41 .4 multiplied by cost 56 .2 degrees.

02:07 On calculating we get minus 23 .030 and the y component of b vector can be written as b sine theta b.

02:21 Substituting the values we get 41 .4 multiplied by sine 56 .2 degrees.

02:31 On calculating, we get 34 .402.

02:36 Since c vector lies in the negative y axis, c vector only has y component which is equal to minus 48 .9 and the x component is 0.

02:55 A vector minus b vector plus c vector can be written as ax minus bx plus cx x cap plus ay minus by plus cy cap...

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