If in a triangle `ABC cos A cos B+sin Asin B sin C = 1`, then the sides are proportional to

Options

A. `1:1:sqrt2`
B. `1:sqrt2:1`
C. `sqrt2:1:1`
D. none

Answer

To solve the problem, we start with the equation given in the triangle \( ABC \):

\[ \cos A \cos B + \sin A \sin B \sin C = 1 \]

### Step 1: Multiply the equation by 2
We first multiply the entire equation by 2:

\[ 2(\cos A \cos B + \sin A \sin B \sin C) = 2 \]

This simplifies to:

\[ 2 \cos A \cos B + 2 \sin A \sin B \sin C = 2 \]

### Step 2: Rewrite the right-hand side
We can express the right-hand side as the sum of squares:

\[ 2 = \sin^2 A + \cos^2 A + \sin^2 B + \cos^2 B \]

### Step 3: Rearranging the equation
Now we rearrange the left-hand side:

\[ 2 \cos A \cos B + 2 \sin A \sin B \sin C = \sin^2 A + \cos^2 A + \sin^2 B + \cos^2 B \]

This leads us to:

\[ \sin^2 A + \sin^2 B - 2 \sin A \sin B + \cos^2 A + \cos^2 B - 2 \cos A \cos B = 0 \]

### Step 4: Grouping terms
We can group the terms:

\[ (\sin A - \sin B)^2 + (\cos A - \cos B)^2 + 2 \sin A \sin B (1 - \sin C) = 0 \]

### Step 5: Analyzing the equation
For the sum of squares to equal zero, each term must be zero:

1. \((\sin A - \sin B)^2 = 0 \Rightarrow \sin A = \sin B\)
2. \((\cos A - \cos B)^2 = 0 \Rightarrow \cos A = \cos B\)
3. \(1 - \sin C = 0 \Rightarrow \sin C = 1\)

From \(\sin A = \sin B\) and \(\cos A = \cos B\), we conclude that:

\[ A = B \]

From \(\sin C = 1\), we find:

\[ C = 90^\circ \]

### Step 6: Finding the angles
Since \(A + B + C = 180^\circ\) and \(C = 90^\circ\):

\[ A + B = 90^\circ \]

Thus, if \(A = B\), then:

\[ 2A = 90^\circ \Rightarrow A = B = 45^\circ \]

### Step 7: Finding the sides' ratios
Now, we know the angles:

- \(A = 45^\circ\)
- \(B = 45^\circ\)
- \(C = 90^\circ\)

Using the sine rule, the sides are proportional to:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]

Substituting the values:

\[ \frac{a}{\sin 45^\circ} = \frac{b}{\sin 45^\circ} = \frac{c}{\sin 90^\circ} \]

This gives us:

\[ \frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{1}{\sqrt{2}}} = \frac{c}{1} \]

Thus, the sides are proportional to:

\[ 1 : 1 : \sqrt{2} \]

### Final Answer
The sides are proportional to \(1 : 1 : \sqrt{2}\).

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