Question

Please show your work!!

Let |a| = 12 at an angle of 25º and |b| = 7 at an angle of 105º. What is the magnitude of a+b? Round to the nearest decimal.
50 points to whoever answers this correctly! The question has no multiple choice answers.

Answer 1

`||\vec a + \vec b||=14.90 \ at \ 52.33 \textdegree`

Explanation:

Given the magnitude of two vectors, "a" and "b," find the magnitude of a+b.

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Here's a step-by-step process to find the magnitude and angle of the vector sum of two given vectors:

(1) - Identify the magnitudes and angles of the two vectors

(2) - Split the vectors into their x and y components. Use trigonometry to find the x and y components of each vector. Round if needed.

(3) - Add the x-components and y-components separately.

(4) - Calculate the magnitude of the vector sum using the Pythagorean theorem. Round if needed.

(5) - Calculate the angle of the vector sum. Round if needed.

`\boxed{\left\begin{array}{ccc}\vec v = < \ v_x, \ v_y > \\\\\text{\underline{Where:}} \\\\ ||\vec v||=√(v_x^2+v_y^2) \\\\ v_x=||\vec v||\cos(\theta)\\\\v_y=||\vec v||\sin(\theta) \\\\ \theta=\tan^(-1)\Big((v_y)/(v_x) \Big) \ (+180\textdegree \ \text{if} \ v_x < 0 )\end{array}\right}`

Note* if the given angles are in degrees, use degrees mode on your calculator.
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Step (1):

`||\vec a|| = 12 \ at \ 25 \textdegree\\\\ ||\vec b|| = 7 \ at \ 105 \textdegree`

Step (2):

Finding vector a:

`\vec a= < ||\vec a||\cos(\theta),||\vec a||\sin(\theta) > \\\\\\\Longrightarrow \vec a= < 12\cos(25\textdegree),12\sin(25\textdegree) > \\\\\\\Longrightarrow \boxed{\vec a= < 10.88,5.07 > }`

Finding vector b:

`\vec b= < ||\vec b||\cos(\theta),||\vec b||\sin(\theta) > \\\\\\\Longrightarrow \vec b= < 7\cos(105\textdegree),7\sin(105\textdegree) > \\\\\\\Longrightarrow \boxed{\vec b= < -1.81,6.76 > }`

Step (3):

`\vec a + \vec b = < a_x+b_x, a_y+b_y > \\\\\\\Longrightarrow \vec a + \vec b= < 10.88+(-1.81),5.07+6.76 > \\\\\\\Longrightarrow \boxed{\vec a + \vec b= < 9.06,11.83 > }`

Step (4):

`||\vec a + \vec b||=√([(\vec a + \vec b)_x]^2+[(\vec a + \vec b)_y]^2) \\\\\\\Longrightarrow ||\vec a + \vec b||=√((9.06)^2+(11.83)^2)\\\\\\\Longrightarrow \boxed`

Step (5):

`\theta=\tan^(-1)\Big(((\vec a + \vec b)_y)/((\vec a + \vec b)_x) \Big)\\\\\\\Longrightarrow \theta=\tan^(-1)\Big((11.83)/(9.06) \Big)\\\\\\\Longrightarrow \boxed{\theta=52.55 \textdegree}`

Thus, the problem is solved.

`||\vec a + \vec b||=14.90 \ at \ 52.33 \textdegree`