The diagonals of a parallelogram are inclined to each other at an angle of `45^@`, while its sides `a and b (a>0)` are inclined to each other at an angle of `30^@`, then the value of `a/b` is

Options

A. `(3)/(2)`
B. `(3+sqrt(5))/(2)`
C. `(3+sqrt(5))/(4)`
D. `(sqrt(5)+1)/(2)`

Explanation

To find the value of \( \frac{a}{b} \) given the conditions of the parallelogram, we will follow these steps:

### Step 1: Understand the given information
We have a parallelogram with:
- Diagonals inclined at an angle of \( 45^\circ \)
- Sides \( a \) and \( b \) inclined at an angle of \( 30^\circ \)

### Step 2: Use the area formula for the parallelogram
The area \( A \) of a parallelogram can be expressed in two ways:
1. Using the diagonals:
\[ A = \frac{1}{2} d_1 d_2 \sin(45^\circ) \]
2. Using the sides:
\[ A = a b \sin(30^\circ) \]

### Step 3: Set the two area expressions equal
Since both expressions represent the same area, we can set them equal to each other:
\[ \frac{1}{2} d_1 d_2 \sin(45^\circ) = a b \sin(30^\circ) \]

### Step 4: Substitute known values
We know:
- \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Substituting these values into the equation gives:
\[ \frac{1}{2} d_1 d_2 \cdot \frac{\sqrt{2}}{2} = a b \cdot \frac{1}{2} \]
This simplifies to:
\[ \frac{\sqrt{2}}{4} d_1 d_2 = \frac{1}{2} ab \]
Multiplying both sides by 4 results in:
\[ \sqrt{2} d_1 d_2 = 2 ab \]
Thus, we have:
\[ d_1 d_2 = \frac{2 ab}{\sqrt{2}} = \sqrt{2} ab \]

### Step 5: Use the cosine rule to express \( d_1 \) and \( d_2 \)
Using the cosine rule for the diagonals:
1. For \( d_1 \):
\[ d_1^2 = a^2 + b^2 - 2ab \cos(150^\circ) \]
Since \( \cos(150^\circ) = -\frac{\sqrt{3}}{2} \):
\[ d_1^2 = a^2 + b^2 + \sqrt{3} ab \]

2. For \( d_2 \):
\[ d_2^2 = a^2 + b^2 - 2ab \cos(30^\circ) \]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ d_2^2 = a^2 + b^2 - \sqrt{3} ab \]

### Step 6: Calculate \( d_1^2 d_2^2 \)
Now we can calculate \( d_1^2 d_2^2 \):
\[ d_1^2 d_2^2 = (a^2 + b^2 + \sqrt{3} ab)(a^2 + b^2 - \sqrt{3} ab) \]
This expands to:
\[ = (a^2 + b^2)^2 - (\sqrt{3} ab)^2 = (a^2 + b^2)^2 - 3a^2b^2 \]

### Step 7: Set the equation for \( d_1 d_2 \)
We know from earlier that:
\[ d_1 d_2 = \sqrt{2} ab \]
Thus:
\[ d_1^2 d_2^2 = 2a^2b^2 \]

### Step 8: Equate the two expressions
Now we equate the two expressions for \( d_1^2 d_2^2 \):
\[ (a^2 + b^2)^2 - 3a^2b^2 = 2a^2b^2 \]
This simplifies to:
\[ (a^2 + b^2)^2 - 5a^2b^2 = 0 \]

### Step 9: Let \( x = \frac{a}{b} \)
Let \( x = \frac{a}{b} \), then \( a^2 = x^2b^2 \):
\[ (x^2b^2 + b^2)^2 - 5x^2b^4 = 0 \]
This simplifies to:
\[ (b^2(x^2 + 1))^2 - 5x^2b^4 = 0 \]
Dividing through by \( b^4 \):
\[ (x^2 + 1)^2 - 5x^2 = 0 \]

### Step 10: Solve the quadratic equation
Expanding gives:
\[ x^4 - 3x^2 + 1 = 0 \]
Let \( y = x^2 \):
\[ y^2 - 3y + 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]
Taking the positive root:
\[ x^2 = \frac{3 + \sqrt{5}}{2} \]
Thus:
\[ x = \sqrt{\frac{3 + \sqrt{5}}{2}} \]

### Final Step: Conclusion
The value of \( \frac{a}{b} \) is:
\[ \frac{a}{b} = \sqrt{\frac{3 + \sqrt{5}}{2}} \]