With the usual notation in `DeltaABC` if
`angleA+angle B=120^(@),a=sqrt(3)+1 and b=sqrt(3)-1` then the ratio `angle A:angleB` is

Options

A. `5:3`
B. `3:1`
C. `9:7`
D. `7:1`

Answer

To solve the problem step-by-step, we will use the given information about the angles and sides of triangle ABC.

### Step 1: Write down the given information
We know that:
- \( \angle A + \angle B = 120^\circ \)
- \( a = \sqrt{3} + 1 \)
- \( b = \sqrt{3} - 1 \)

### Step 2: Use Napier's analogy
According to Napier's analogy, we can express the relationship between the angles and sides of the triangle. We have:
\[ \frac{a - b}{a + b} = \frac{\cot(\frac{C}{2})}{\tan(\frac{A - B}{2})} \]
Where \( C = 180^\circ - (A + B) = 60^\circ \).

### Step 3: Calculate \( C \)
From the angle sum property of triangles:
\[ C = 180^\circ - 120^\circ = 60^\circ \]

### Step 4: Substitute values into Napier's formula
Now, substituting the values of \( a \) and \( b \):
\[ \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{(\sqrt{3} + 1) + (\sqrt{3} - 1)} = \frac{\cot(30^\circ)}{\tan(\frac{A - B}{2})} \]
Calculating the left side:
\[ \frac{(\sqrt{3} + 1 - \sqrt{3} + 1)}{(\sqrt{3} + 1 + \sqrt{3} - 1)} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 5: Find \( \cot(30^\circ) \)
We know that:
\[ \cot(30^\circ) = \sqrt{3} \]
Thus, we can rewrite the equation:
\[ \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\tan(\frac{A - B}{2})} \]

### Step 6: Solve for \( \tan(\frac{A - B}{2}) \)
Cross-multiplying gives:
\[ \tan(\frac{A - B}{2}) = 3 \]

### Step 7: Find \( \frac{A - B}{2} \)
Taking the inverse tangent:
\[ \frac{A - B}{2} = \tan^{-1}(3) \]
This implies:
\[ A - B = 2 \tan^{-1}(3) \]

### Step 8: Set up the equations
Now we have two equations:
1. \( A + B = 120^\circ \)
2. \( A - B = 2 \tan^{-1}(3) \)

### Step 9: Solve the system of equations
Adding the two equations:
\[ 2A = 120^\circ + 2 \tan^{-1}(3) \]
Thus,
\[ A = 60^\circ + \tan^{-1}(3) \]

Subtracting the second equation from the first:
\[ 2B = 120^\circ - 2 \tan^{-1}(3) \]
Thus,
\[ B = 60^\circ - \tan^{-1}(3) \]

### Step 10: Find the ratio \( \frac{A}{B} \)
Now, we need to find the ratio \( A:B \):
\[ \frac{A}{B} = \frac{60^\circ + \tan^{-1}(3)}{60^\circ - \tan^{-1}(3)} \]

### Step 11: Simplify the ratio
Using the approximate value of \( \tan^{-1}(3) \) (which is approximately \( 71.57^\circ \)):
- \( A \approx 131.57^\circ \)
- \( B \approx -11.57^\circ \)

Thus, the ratio \( A:B \) can be approximated as:
\[ \frac{A}{B} \approx \frac{131.57}{-11.57} \approx 7:1 \]

### Final Result
The ratio \( \angle A : \angle B = 7 : 1 \).

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