How do you get the 112.6 angle in this problem?

Question

Hi all could some one help understand this problem? I understand that the new resulting vector after the wind is -200.5 i + 482.5 J.

and if you want to calculate the magnitude you obtain 522.5, but how on earth do you obtain 112.6? and why do you multiply the magnitude by the sum of the magnitude of the vector divided by the magnitude?

Answer 1

to get the angle they use sin inverse and cosine inverse of the fractions next to I and k respectively in the very last line of the equation.

The way to do this is to think "what angle has $$\cos{\theta}=\frac{-200.5}{522.5}?$$

The answer can be gotten using a calculator and using the inverse cos button, it looks like cos but with a negative 1 exponent. It can also look like $$ \arccos{\theta} $$

they mean the same thing. You simply use the button and type the fraction. make sure that the answer has the sin you need as well.

For the second part, remember that a vector is direction and magnitude, but there is always the vector with magnitude one. Dividing both the I and K magnitudes with the main magnitude gives you the numbers for this "magnitude 1" vector, and because its easy to get to any other magnitude using this vector with magnitude one, we usually want to calculate it.

Answer 2

$112.6^\circ$ is the angle $\theta$ such that $\cos{\theta}=\frac{-200.5}{522.5}$ and $\sin{\theta}=\frac{482.5}{522.5}$.

There are several methods to solve this, all using inverse trigonometric functions.

For example, taking the arctangent of $\frac{\sin\theta}{\cos\theta}$ gives us a result of $-67.4$ (in degrees). We can't take this, since its cosine is supposed to be negative yet its cosine is positive.

We then use the fact that solutions to $\tan\theta=c$ for constant $c$ is in the form $a+180k$ where $a$ is a real value and $k$ is an integer. This fact allows us to look at $-67.4+180=112.6$ as the potential argument for the vector, which when verified, works.

We divide the net force vector by its magnitude to get its unit vector. This unit vector is in the same direction as the original vector and its magnitude of $1$, which allows us to conveniently set $\cos{\theta}$ to the i-component and $\sin{\theta}$ to the j-component.

Answer 3

If you have a vector of the form: $a(\cos \theta \vec i + \sin\theta \vec j)$, where $a\ge0$, then it means the magnitude of vector is $a$ and the vector makes $\theta$ with positive direction of $x$-axis.
Note that in this the $i$ component and $j$ component should square and add to unity. That is the vector should be a unit vector.
So in general if you have a vector:$a\vec i+b\vec j$ then you can always convert it into form: $$ \sqrt{a^2+b^2} \left( \frac a {\sqrt{a^2+b^2}} \vec i+\frac b {\sqrt{a^2+b^2}} \vec j \right)~,$$ and can find a $\theta$ such that $$\sin\theta =\frac a {\sqrt{a^2+b^2}} \quad\text{and} \quad\cos \theta=\frac b {\sqrt{a^2+b^2}}~.$$ Now, you can see that the magnitude of vector is ${\sqrt{a^2+b^2}}$ and vector makes $\theta$ with the positive $x$-axis.