The two vectors A and B in the figure have equal magnitudes of 13.5 m and the angles are θ1 = 33° and θ2 = 110°. How to find (a) the x component and (b) the y component of their vector sum R , (c) the magnitude of R, and (d) the angle R ?

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The two vectors A and B in the figure have equal magnitudes of 13.5 m and the angles are θ1 = 33° and θ2 = 110°. How to find (a) the x component and (b) the y component of their vector sum R , (c) the magnitude of R, and (d) the angle R ?

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Answer 1 · 2015-09-22T18:25:10

Here's what I got.

Explanation:

enter image source here

I don't wave a good way of drawing you a diagram, so I'll try to walk you through the steps as they come along.

So, the idea here is that you can find the $x$ $x$ -component and the $y$ $y$ -component of the vector sum, $R$ $R$ , by adding the $x$ $x$ -components and $y$ $y$ -components, respectively, of the $\to a$ $vec(a)$ and $\to b$ $vec(b)$ vectors.

For vector $\to a$ $vec(a)$ , things are pretty straighforward. The $x$ $x$ -component will be the projection of the vector on the $x$ $x$ -axis, which is equal to

$a_{x} = a \cdot cos (θ_{1})$ $a_x = a * cos(theta_1)$

Likewise, the $y$ $y$ -component will be the projection of the vector on the $y$ $y$ -axis

$a_{y} = a \cdot sin (θ_{1})$ $a_y = a * sin(theta_1)$

For vector $\to b$ $vec(b)$ , things are a little more complicated. More specifically, finding the corresponding angles will be a little tricky.

The angle between $\to a$ $vec(a)$ and $\to b$ $vec(b)$ is

$θ_{3} = 180^{\circ} - θ_{2} = 180^{\circ} - 110^{\circ} = 70^{\circ}$ $theta_3 = 180^@ - theta_2 = 180^@ - 110^@ = 70^@$

Draw a parallel line to the $x$ $x$ -axis that intersects the point where the tail of $\to b$ $vec(b)$ and head of $\to a$ $vec(a)$ meet.

![http://www.regentsprep.org/regents/math/geometry/multiplechoicereviewg/http://quadrilaterals.htm](https://useruploads.socratic.org/cLDEBvYGT9W8CAVtev1c_QuadPic5.jpg)

In your case, line $m$ $m$ will be the $x$ $x$ -axis and line $a$ $a$ the parallel line you draw.

In this drawing, $∠ 6$ $angle6$ is $θ_{1}$ $theta_1$ . You know that $∠ 6$ $angle6$ is equal to $∠ 3$ $angle3$ , $∠ 2$ $angle2$ , and $∠ 7$ $angle7$ .

The angle between $\to b$ $vec(b)$ and the $x$ $x$ -axis will be equal to

$180^{\circ} - (θ_{1} + θ_{2}) = 180^{\circ} - 143^{\circ} = 37^{\circ}$ $180^@ - (theta_1 + theta_2) = 180^@ - 143^@ = 37^@$

This means that the $x$ $x$ -component of vector $\to b$ $vec(b)$ will be

$b_{x} = b \cdot cos (37^{\circ})$ $b_x = b * cos(37^@)$

Now, because the angle between the $x$ $x$ -component and the $y$ $y$ -component of a vector is equal to $90^{\circ}$ $90^@$ , it follows that the angle for the $y$ $y$ -component of $\to b$ $vec(b)$ will be

$90^{\circ} - 37^{\circ} = 53^{\circ}$ $90^@ - 37^@ = 53^@$

The $y$ $y$ -component will thus be

$b_{y} = b \cdot sin (37^{\circ})$ $b_y = b * sin(37^@)$

Now, keep in mind that the $x$ $x$ -component of $\to b$ $vec(b)$ is oriented in the opposite direction of the $x$ $x$ -component of $\to a$ $vec(a)$ . This means that the $x$ $x$ -component of $\to R$ $vec(R)$ will be

$R_{x} = a_{x} + b_{x}$ $R_x = a_x + b_x$

$R_{x} = 13.5 \cdot cos (33^{\circ}) - 13.5 \cdot cos (37^{\circ})$ $R_x = 13.5 * cos(33^@) - 13.5 * cos(37^@)$

$R_{x} = 13.5 \cdot 0.04 = 0.54 m$ $R_x = 13.5 * 0.04 = color(green)("0.54 m")$

The $y$ $y$ -components are oriented in the same direction, so you have

$R_{y} = a_{y} + b_{y}$ $R_y = a_y + b_y$

$R_{y} = 13.5 \cdot [sin (110^{\circ}) + sin (37^{\circ})]$ $R_y = 13.5 * [sin(110^@) + sin(37^@)]$

$R_{y} = 13.5 \cdot 1.542 = 20.82 m$ $R_y = 13.5 * 1.542 = color(green)("20.82 m")$

The magnitude of $\to R$ $vec(R)$ will be

$R^{2} = R_{x}^{2} + R_{y}^{2}$ $R^2 = R_x^2 + R_y^2$

$R = \sqrt{0.54^{2} + 20.82^{2}} m = 20.83 m$ $R = sqrt(0.54""^2 + 20.82""^2)" m" = color(green)("20.83 m")$

To get the angle of $\to R$ $vec(R)$ , simply use

$tan (θ_{R}) = \frac{R_{y}}{R_{x}} \Rightarrow θ_{R} = arctan (\frac{R_{y}}{R_{x}})$ $tan(theta_R) = R_y/R_x implies theta_R = arctan(R_y/R_x)$

$theta_R = arctan((20.82color(red)(cancel(color(black)("m"))))/(0.54color(red)(cancel(color(black)("m"))))) = color(green)(88.6""^@)$

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The two vectors A and B in the figure have equal magnitudes of 13.5 m and the angles are θ1 = 33° and θ2 = 110°. How to find (a) the x component and (b) the y component of their vector sum R , (c) the magnitude of R, and (d) the angle R ?

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