A right triangle has perimeter of length 7 and hypotenuse of length `3.` If `theta` is the larger non-right angle in the triangle, then the value of `costhetae q u a ldot` `(sqrt(6)-sqrt(2))/4` (b) `(4+sqrt(2))/6` `(4-sqrt(2))/3` (d) `(4-sqrt(2))/6`

Options

A. `(sqrt6-sqrt2)/4`
B. `(4+sqrt2)/6`
C. `(4-sqrt2)/6`
D. `(4-sqrt2)/6`

Explanation

To solve the problem, we need to find the value of \( \cos \theta \) for a right triangle with a perimeter of 7 and a hypotenuse of length 3. Here are the steps to solve the problem:

### Step 1: Set up the equations
Let the lengths of the two legs of the triangle be \( a \) and \( b \). The hypotenuse \( c \) is given as 3. The perimeter of the triangle is given as 7. Therefore, we can write the following equations:

1. \( a + b + c = 7 \)
2. \( a^2 + b^2 = c^2 \)

Substituting \( c = 3 \) into the perimeter equation gives us:
\[ a + b + 3 = 7 \implies a + b = 4 \]

### Step 2: Express one variable in terms of the other
From the equation \( a + b = 4 \), we can express \( a \) in terms of \( b \):
\[ a = 4 - b \]

### Step 3: Substitute into the Pythagorean theorem
Now, substitute \( a = 4 - b \) into the Pythagorean theorem:
\[ (4 - b)^2 + b^2 = 3^2 \]
Expanding this gives:
\[ (16 - 8b + b^2) + b^2 = 9 \]
Combining like terms results in:
\[ 2b^2 - 8b + 16 - 9 = 0 \]
This simplifies to:
\[ 2b^2 - 8b + 7 = 0 \]

### Step 4: Solve the quadratic equation
Now we can use the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) where \( A = 2, B = -8, C = 7 \):
\[ b = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 2 \cdot 7}}{2 \cdot 2} \]
Calculating the discriminant:
\[ b = \frac{8 \pm \sqrt{64 - 56}}{4} = \frac{8 \pm \sqrt{8}}{4} = \frac{8 \pm 2\sqrt{2}}{4} \]
This simplifies to:
\[ b = 2 \pm \frac{\sqrt{2}}{2} \]

### Step 5: Determine the lengths of the sides
Since \( b \) must be the smaller leg, we take:
\[ b = 2 - \frac{\sqrt{2}}{2} \]
Then, substituting back to find \( a \):
\[ a = 4 - b = 4 - \left(2 - \frac{\sqrt{2}}{2}\right) = 2 + \frac{\sqrt{2}}{2} \]

### Step 6: Calculate \( \cos \theta \)
Since \( \theta \) is the larger angle, we use the definition of cosine:
\[ \cos \theta = \frac{b}{c} = \frac{2 - \frac{\sqrt{2}}{2}}{3} \]
To simplify:
\[ \cos \theta = \frac{4 - \sqrt{2}}{6} \]

### Conclusion
Thus, the value of \( \cos \theta \) is:
\[ \cos \theta = \frac{4 - \sqrt{2}}{6} \]