In triangle ABC, `a=3, b=4, c=2`. Point D and E trisect the side BC. If `angleDAE=theta`, then `cot^(2)theta` is divisible by :

Options

A. 2
B. 3
C. 5
D. 7

Answer

To solve the problem step by step, we will analyze triangle ABC with the given sides and points D and E that trisect side BC.

### Step 1: Identify the triangle and its sides
Given:
- \( a = 3 \) (length of side BC)
- \( b = 4 \) (length of side AC)
- \( c = 2 \) (length of side AB)

### Step 2: Trisect side BC
Since D and E trisect side BC, we can divide BC into three equal parts:
- \( BD = DE = EC = 1 \)

### Step 3: Use the Cosine Rule in triangle ABC
We will use the Cosine Rule to find the angles of triangle ABC. The Cosine Rule states:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting the values:
\[ \cos B = \frac{3^2 + 2^2 - 4^2}{2 \cdot 3 \cdot 2} = \frac{9 + 4 - 16}{12} = \frac{-3}{12} = -\frac{1}{4} \]

### Step 4: Use the Cosine Rule in triangle ABD
Now, we will apply the Cosine Rule in triangle ABD:
\[ \cos B = \frac{AD^2 + BD^2 - AB^2}{2 \cdot AD \cdot BD} \]
Let \( AD = x \):
\[ -\frac{1}{4} = \frac{x^2 + 1^2 - 2^2}{2 \cdot x \cdot 1} \]
This simplifies to:
\[ -\frac{1}{4} = \frac{x^2 + 1 - 4}{2x} \]
Cross-multiplying gives:
\[ -x^2 - 3 = -8x \implies x^2 - 8x + 3 = 0 \]
Using the quadratic formula:
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 - 12}}{2} = \frac{8 \pm \sqrt{52}}{2} = \frac{8 \pm 2\sqrt{13}}{2} = 4 \pm \sqrt{13} \]
Taking the positive root:
\[ AD = 4 - \sqrt{13} \]

### Step 5: Use the Cosine Rule in triangle ABE
Now we will apply the Cosine Rule in triangle ABE:
\[ \cos C = \frac{AB^2 + AE^2 - BE^2}{2 \cdot AB \cdot AE} \]
Let \( AE = y \):
\[ \cos C = \frac{2^2 + y^2 - 1^2}{2 \cdot 2 \cdot y} \]
We already found \( \cos C \) using triangle ABC:
\[ \cos C = \frac{3^2 + 4^2 - 2^2}{2 \cdot 3 \cdot 4} = \frac{9 + 16 - 4}{24} = \frac{21}{24} = \frac{7}{8} \]
Setting the two expressions equal:
\[ \frac{7}{8} = \frac{4 + y^2 - 1}{4y} \]
Cross-multiplying:
\[ 7 \cdot 4y = 8(3 + y^2) \implies 28y = 24 + 8y^2 \]
Rearranging gives:
\[ 8y^2 - 28y + 24 = 0 \]
Using the quadratic formula:
\[ y = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 8 \cdot 24}}{2 \cdot 8} = \frac{28 \pm \sqrt{784 - 768}}{16} = \frac{28 \pm 4}{16} \]
Calculating gives:
\[ y = \frac{32}{16} = 2 \quad \text{or} \quad y = \frac{24}{16} = \frac{3}{2} \]
Taking the positive root:
\[ AE = 2 \]

### Step 6: Calculate cotangent of angle DAE
In triangle ADE, we have:
- \( DE = 1 \)
- \( AD = 4 - \sqrt{13} \)
- \( AE = 2 \)

Using the Cosine Rule again:
\[ \cos \theta = \frac{AD^2 + AE^2 - DE^2}{2 \cdot AD \cdot AE} \]
Substituting the values:
\[ \cos \theta = \frac{(4 - \sqrt{13})^2 + 2^2 - 1^2}{2 \cdot (4 - \sqrt{13}) \cdot 2} \]
Calculating gives:
\[ \cos \theta = \frac{(16 - 8\sqrt{13} + 13 + 4 - 1)}{4(4 - \sqrt{13})} = \frac{32 - 8\sqrt{13}}{4(4 - \sqrt{13})} \]
Simplifying gives:
\[ \cos \theta = \frac{8 - 2\sqrt{13}}{4 - \sqrt{13}} \]

### Step 7: Find cot^2(theta)
Using the identity \( \cot^2 \theta = \frac{1 - \cos^2 \theta}{\cos^2 \theta} \), we can find \( \cot^2 \theta \).

### Step 8: Check divisibility
Finally, we check which of the options (2, 3, 5, 7) divide \( \cot^2 \theta \).