ABC is a triangle is which AB = 6 cm, BC = 8 cm and CA = 10 cm. What is the value of cot(A/4) ?

Options

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

Explanation

To find the value of \( \cot\left(\frac{A}{4}\right) \) for triangle ABC where \( AB = 6 \, \text{cm}, BC = 8 \, \text{cm}, \) and \( CA = 10 \, \text{cm} \), we can follow these steps:

### Step 1: Calculate the semi-perimeter \( s \) of triangle ABC.
The semi-perimeter \( s \) is given by the formula:
\[ s = \frac{AB + BC + CA}{2} \]
Substituting the values:
\[ s = \frac{6 + 8 + 10}{2} = \frac{24}{2} = 12 \, \text{cm} \]

**Hint:** The semi-perimeter is half the sum of all sides of the triangle.

### Step 2: Use the formula for \( \tan\left(\frac{A}{2}\right) \).
The formula for \( \tan\left(\frac{A}{2}\right) \) is:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} \]
Where \( a = BC, b = CA, c = AB \). Thus:
- \( a = 8 \)
- \( b = 10 \)
- \( c = 6 \)

Substituting the values:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(12 - 10)(12 - 6)}{12(12 - 8)}} \]
Calculating each term:
\[ = \sqrt{\frac{(2)(6)}{12(4)}} = \sqrt{\frac{12}{48}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

**Hint:** Remember that \( s - a, s - b, \) and \( s - c \) represent the lengths of the triangle sides subtracted from the semi-perimeter.

### Step 3: Calculate \( \cot\left(\frac{A}{2}\right) \).
Since \( \cot\left(\frac{A}{2}\right) = \frac{1}{\tan\left(\frac{A}{2}\right)} \):
\[ \cot\left(\frac{A}{2}\right) = \frac{1}{\frac{1}{2}} = 2 \]

**Hint:** The cotangent is the reciprocal of the tangent.

### Step 4: Use the formula for \( \cot\left(\frac{A}{4}\right) \).
The relationship between \( \cot\left(\frac{A}{4}\right) \) and \( \cot\left(\frac{A}{2}\right) \) is given by:
\[ \cot\left(\frac{A}{2}\right) = \frac{\cot^2\left(\frac{A}{4}\right) - 1}{2 \cot\left(\frac{A}{4}\right)} \]
Let \( x = \cot\left(\frac{A}{4}\right) \). Then we have:
\[ 2 = \frac{x^2 - 1}{2x} \]
Cross-multiplying gives:
\[ 4x = x^2 - 1 \]
Rearranging the equation:
\[ x^2 - 4x - 1 = 0 \]

**Hint:** This is a quadratic equation that can be solved using the quadratic formula.

### Step 5: Solve the quadratic equation.
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -4, c = -1 \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5} \]

**Hint:** The solutions represent two possible values for \( \cot\left(\frac{A}{4}\right) \).

### Step 6: Determine the correct value.
Since \( A \) is an angle in a triangle, \( \cot\left(\frac{A}{4}\right) \) must be positive. Thus, we take:
\[ \cot\left(\frac{A}{4}\right) = 2 + \sqrt{5} \]

### Final Answer:
\[ \cot\left(\frac{A}{4}\right) = 2 + \sqrt{5} \]