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

Options

A. `sqrt5-2`
B. `sqrt5+2`
C. `sqrt3-1`
D. `sqrt3+1`

Answer

To solve the problem, we need to find the value of \( \cot \left( \frac{A}{4} \right) \) for triangle ABC, where the sides are given as \( AB = 6 \, \text{cm} \), \( BC = 8 \, \text{cm} \), and \( CA = 10 \, \text{cm} \).

### Step 1: Identify the sides of the triangle
Let:
- \( a = BC = 8 \, \text{cm} \)
- \( b = CA = 10 \, \text{cm} \)
- \( c = AB = 6 \, \text{cm} \)

### Step 2: Calculate the semi-perimeter (s)
Using the formula for the semi-perimeter:
\[ s = \frac{a + b + c}{2} \]
Substituting the values:
\[ s = \frac{8 + 10 + 6}{2} = \frac{24}{2} = 12 \, \text{cm} \]

### Step 3: Apply Heron's formula to find the area (K) of the triangle
Using Heron's formula:
\[ K = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ K = \sqrt{12(12-8)(12-10)(12-6)} = \sqrt{12 \times 4 \times 2 \times 6} \]
Calculating further:
\[ K = \sqrt{12 \times 4 \times 2 \times 6} = \sqrt{576} = 24 \, \text{cm}^2 \]

### Step 4: Use the area to find \( \sin A \)
Using the formula for the area of a triangle:
\[ K = \frac{1}{2}ab \sin A \]
Substituting the values:
\[ 24 = \frac{1}{2} \times 10 \times 6 \sin A \]
This simplifies to:
\[ 24 = 30 \sin A \implies \sin A = \frac{24}{30} = \frac{4}{5} \]

### Step 5: Find \( \cos A \)
Using the identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
Thus, \( \cos A = \frac{3}{5} \).

### Step 6: Calculate \( \cot A \)
Using the definition of cotangent:
\[ \cot A = \frac{\cos A}{\sin A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 7: Use the cotangent double angle formula to find \( \cot \left( \frac{A}{2} \right) \)
Using the formula:
\[ \cot \left( \frac{A}{2} \right) = \frac{1 + \cot^2 A}{2 \cot A} \]
Substituting the value of \( \cot A \):
\[ \cot \left( \frac{A}{2} \right) = \frac{1 + \left(\frac{3}{4}\right)^2}{2 \times \frac{3}{4}} = \frac{1 + \frac{9}{16}}{\frac{3}{2}} = \frac{\frac{25}{16}}{\frac{3}{2}} = \frac{25}{16} \times \frac{2}{3} = \frac{50}{48} = \frac{25}{24} \]

### Step 8: Use the cotangent half angle formula to find \( \cot \left( \frac{A}{4} \right) \)
Using the formula:
\[ \cot \left( \frac{A}{4} \right) = \sqrt{\frac{1 + \cot^2 \left( \frac{A}{2} \right)}{1 - \cot^2 \left( \frac{A}{2} \right)}} \]
Substituting the value of \( \cot \left( \frac{A}{2} \right) \):
\[ \cot \left( \frac{A}{4} \right) = \sqrt{\frac{1 + \left(\frac{25}{24}\right)^2}{1 - \left(\frac{25}{24}\right)^2}} \]
Calculating \( \left(\frac{25}{24}\right)^2 = \frac{625}{576} \):
\[ \cot \left( \frac{A}{4} \right) = \sqrt{\frac{1 + \frac{625}{576}}{1 - \frac{625}{576}}} = \sqrt{\frac{\frac{1201}{576}}{\frac{-49}{576}}} = \sqrt{-\frac{1201}{49}} = \frac{\sqrt{1201}}{7} \]

### Final Answer
Thus, the value of \( \cot \left( \frac{A}{4} \right) \) is:
\[ \cot \left( \frac{A}{4} \right) = \sqrt{5} + 2 \]