Two vectors, A and B, are added together to form the vector `C = A + B.` The realtionship between the magnitudes of these vectors is given by:
`C _(x) =0`
` C _(y) = A sin 60^(@) + B sin 30^(@)`
`A _(x) and A _(y)` point in the positive x and y directions, respectively.
How does the magnitude of A compare with that of B ?

Options

A. `A =B`
B. `A =1.7B`
C. `A =0.4B`
D. `A = 0.5B`

Answer

To solve the problem, we need to analyze the given information about the vectors A and B and their resultant vector C. We will follow these steps:

### Step 1: Understand the components of the vectors
Given that:
- \( C_x = 0 \)
- \( C_y = A \sin(60^\circ) + B \sin(30^\circ) \)

We know that the x-component of vector C is zero, which implies that the x-components of vectors A and B must cancel each other out.

### Step 2: Set up the x-component equation
From the information provided, we can express the x-components of vectors A and B:
- \( A_x = A \cos(60^\circ) \)
- \( B_x = B \cos(30^\circ) \)

Since \( C_x = 0 \), we have:
\[ A_x + B_x = 0 \]
This leads to:
\[ A \cos(60^\circ) + B \cos(30^\circ) = 0 \]

### Step 3: Substitute the values of cosine
Using the known values:
- \( \cos(60^\circ) = \frac{1}{2} \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)

Substituting these values into the equation gives:
\[ A \left(\frac{1}{2}\right) + B \left(\frac{\sqrt{3}}{2}\right) = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation yields:
\[ A \cdot \frac{1}{2} = -B \cdot \frac{\sqrt{3}}{2} \]

### Step 5: Isolate A in terms of B
Multiplying both sides by 2 to eliminate the fraction:
\[ A = -B \sqrt{3} \]

Since magnitudes are always positive, we can drop the negative sign:
\[ A = B \sqrt{3} \]

### Step 6: Conclusion
This shows that the magnitude of vector A is greater than the magnitude of vector B by a factor of \( \sqrt{3} \). Therefore, we can conclude:
\[ A > B \]