EFGH is a rhombus such that the angle EFG is `60^(@)`. The magnitude of vectors ` bar(FH) ` and {m `bar(EG)`} are equal where m is a scalar. What is the value of m?

Options

A. 3
B. 1.5
C. `sqrt(2)`
D. `sqrt(3)`

Explanation

To solve the problem, we need to find the value of the scalar \( m \) such that the magnitudes of the vectors \( \overline{FH} \) and \( m \overline{EG} \) are equal in the rhombus \( EFGH \) with the angle \( \angle EFG = 60^\circ \).

### Step-by-Step Solution:

1. **Understanding the Rhombus Properties**:
- In a rhombus, opposite angles are equal, and the diagonals bisect each other at right angles. Given \( \angle EFG = 60^\circ \), we can deduce the other angles:
- \( \angle EGF = 60^\circ \)
- \( \angle EHF = 120^\circ \)
- \( \angle HGE = 120^\circ \)

2. **Labeling the Diagonals**:
- Let the diagonals \( \overline{EG} \) and \( \overline{FH} \) intersect at point \( O \).
- Let \( EO = A \) and \( FO = B \). Since the diagonals bisect each other, we have:
- \( OG = A \)
- \( OH = B \)

3. **Using Triangle Properties**:
- In triangle \( EFO \), we can use the tangent of angle \( \angle EFO \):
\[ \tan(30^\circ) = \frac{EO}{FO} = \frac{A}{B} \]
- Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{A}{B} = \frac{1}{\sqrt{3}} \implies A = \frac{B}{\sqrt{3}} \implies B = \sqrt{3}A \]

4. **Finding the Lengths of the Diagonals**:
- The length of diagonal \( \overline{FH} \) is:
\[ FH = 2B \]
- The length of diagonal \( \overline{EG} \) is:
\[ EG = 2A \]

5. **Relating the Lengths**:
- From the previous steps, we substitute \( B \):
\[ FH = 2B = 2(\sqrt{3}A) = 2\sqrt{3}A \]
- And we already have \( EG = 2A \).

6. **Setting Up the Equation**:
- According to the problem, the magnitudes are equal:
\[ |FH| = m |EG| \]
- Substituting the lengths we found:
\[ 2\sqrt{3}A = m(2A) \]
- Dividing both sides by \( 2A \) (assuming \( A \neq 0 \)):
\[ \sqrt{3} = m \]

7. **Conclusion**:
- Therefore, the value of \( m \) is:
\[ m = \sqrt{3} \]