Question

You are given two vectors A⃗ =−3.00i^+7.00j^ and B⃗ =5.00i^+2.00j^. Let counterclockwise angles be positive.
1- What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
2-What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
3-Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees

Answer 1

Answer:

Ax = -3       By= 7         tan phi =  -7 / 3    phi = -66.8

Since theta is in the second quadrant   theta = 180 - 66.8 = 113.2 deg

Bx = 5        By = 2        tan theta = .4        theta = 21.8 deg

Cx = Ax + Bx = 2       Cy = Ay + By = 9

tan theta = 9 / 2 = 4.5      theta = 77.5 deg

Answer 2

1) The solution is $\theta_{1} \approx 113.141^{\circ}$.

2) The solution is $\theta_{1} \approx 21.874^{\circ}$.

3) The solution is $\theta_{1} \approx 77.467^{\circ}$.

1) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

$\cos \theta = \frac{\vec A\,\bullet \, \vec u}{\|\vec A\|\cdot \|\vec u\|}$ (1)

Where $\|\vec A\|$ and $\|\vec u\|$ are the norms of $\vec A$ and $\vec u$.

If we know that $\vec A = (-3, 7)$ and $\vec u = (1, 0)$, then the angle that $\vec A$ make with the x-axis is:

$\|\vec A\| = \sqrt{(-3)^{2}+7^{2}}$

$\|\vec A\| = \sqrt{58}$

$\|\vec u\| = 1$

$\cos \theta = \frac{(-3)\cdot (1)}{\sqrt{58} \cdot 1}$

$\cos \theta \approx -0.393$

The solution is $\theta_{1} \approx 113.141^{\circ}$.

2) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

$\cos \theta = \frac{\vec B\,\bullet \, \vec u}{\|\vec B\|\cdot \|\vec u\|}$ (1)

Where $\|\vec B\|$ and $\|\vec u\|$ are the norms of $\vec B$ and $\vec u$.

If we know that $\vec B = (5, 2)$ and $\vec u = (1, 0)$, then the angle that $\vec A$ make with the x-axis is:

$\|\vec B\| = \sqrt{5^{2}+2^{2}}$

$\|\vec B\| = \sqrt{29}$

$\|\vec u\| = 1$

$\cos \theta = \frac{(5)\cdot (1)}{\sqrt{29} \cdot 1}$

$\cos \theta \approx 0.928$

The solution is $\theta_{1} \approx 21.874^{\circ}$.

3) In this case, we need to find the value of $\vec C$ by vectorial sum before calculating the angles with respect to x-axis:

$\vec C = \vec A + \vec B = (-3, 7) + (5, 2)$

$\vec C = (2, 9)$

In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

$\cos \theta = \frac{\vec C\,\bullet \, \vec u}{\|\vec C\|\cdot \|\vec u\|}$ (1)

Where $\|\vec C\|$ and $\|\vec u\|$ are the norms of $\vec C$ and $\vec u$.

If we know that $\vec B = (2, 9)$ and $\vec u = (1, 0)$, then the angle that $\vec A$ make with the x-axis is:

$\|\vec C\| = \sqrt{2^{2}+9^{2}}$

$\|\vec C\| = \sqrt{85}$

$\|\vec u\| = 1$

$\cos \theta = \frac{(2)\cdot (1)}{\sqrt{85} \cdot 1}$

$\cos \theta \approx 0.217$

The solution is $\theta_{1} \approx 77.467^{\circ}$.

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