Question

Question 3(Multiple Choice Worth 2 points)(07.02 MC)Which of the following represents vector w = -45i + 28j in trigonometric form? Round 0 to the nearest degree.Ow=53cos 32°i + 53sin 32°jOw= 53c0s 148°i + 53sin 148°jOw=35sin 32°i + 35cos 32°jOw= 35sin 148°İ + 35c0s 148°)Question 4(Multiple Choice Worth 2 points)

Answer 1

The vector w = -45i + 28j in trigonometric form, rounded to the nearest degree, is w = 53cos(148°)i + 53sin(148°)j, representing the vector's magnitude and direction.

Step-by-step explanation:

The student's question revolves around converting a vector given in Cartesian form into its trigonometric form. In order to find the trigonometric form of the vector w = -45i + 28j, we need to calculate both the magnitude of the vector and the angle it makes with the positive x-axis. The magnitude can be found using the Pythagorean theorem:

√((-45)^2 + (28)^2) = √(2025 + 784) = √(2809) = 53.

Next, we calculate the angle θ using the arc tangent function based on the components of the vector:

θ = atan2(28, -45) ≈ atan2(j, i) = atan2(opposite, adjacent).

Since the vector is in the second quadrant due to the negative i (x) component and positive j (y) component, we calculate the angle with respect to the positive x-axis:

θ = 180° + tan¹(28 / -45) ≈ 180° - tan¹(28 / 45) ≈ 148° (rounded to the nearest degree).

Now we can write vector w in trigonometric form:

w = 53cos(148°)i + 53sin(148°)j.

Therefore, the trigonometric form that represents vector w is w = 53cos(148°)i + 53sin(148°)j.

Answer 2

Step 1. The vector that we have is:

`\vec{w}=-45i+28j`

This vector is in the form

`\vec{w}=ai+bj`

where

`\begin{gathered} a=-45 \\ b=28 \end{gathered}`

And we need to find the trigonometric form of the vector.

Step 2. To find the trigonometric form we use this formula:

`\vec{w}=|w|cos\theta i+|w|sin\theta j`

Where |w| is the length of the vector and theta is the angle.

Step 3. Finding |w|, the length of the vector:

`|w|=√(a^2+b^2)`

in this case:

`|w|=√((-45)^2+28^2)`

Solving the operations the result is:

`\begin{gathered} \lvert w\rvert=√(2025+284) \\ |w|=53 \end{gathered}`

Step 4. Now, we find the angle using:

`\theta=\tan^(-1)((b)/(a))`

In this case:

`\theta=\tan^(-1)((28)/(-45))`

Solving the operations and rounding to the nearest tenth of a degree:

`\theta=-32`

Step 5. Going back to the formula from step 2:

`\vec{w}=\lvert w\rvert cos\theta\imaginaryI+\lvert w\rvert s\imaginaryI n\theta j`

Substituting |w| and theta:

`\vec{w}=53cos(-32)\imaginaryI+53s\imaginaryI n(-32)j`

We can either leave the result as it is, or use the following properties of the cosine and sine to simplify:

`\begin{gathered} cos(-A)=cosA \\ sin(-A)=-sinA \end{gathered}`

And the result is simplified as follows:

`\vec{w}=53cos(32)\imaginaryI-53s\imaginaryI n(32)j`

Answer:

`\vec{w}=53cos(32)\imaginaryI-53s\imaginaryI n(32)j`