Consider a `DeltaABC` in which the sides are `a = (n +1), b = (n + 1), c = n` with `tan C = 4//3`, then the value of `Delta` is _____

Explanation

To find the area of triangle ABC with sides \( a = n + 1 \), \( b = n + 2 \), and \( c = n \) given that \( \tan C = \frac{4}{3} \), we can follow these steps:

### Step 1: Calculate \( \cos C \)
Since \( \tan C = \frac{4}{3} \), we can find \( \cos C \) using the identity:
\[ \cos C = \frac{1}{\sqrt{1 + \tan^2 C}} = \frac{1}{\sqrt{1 + \left(\frac{4}{3}\right)^2}} = \frac{1}{\sqrt{1 + \frac{16}{9}}} = \frac{1}{\sqrt{\frac{25}{9}}} = \frac{3}{5} \]

**Hint:** Use the Pythagorean identity to find \( \cos C \) from \( \tan C \).

### Step 2: Use the Law of Cosines
Using the Law of Cosines:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Substituting \( a = n + 1 \), \( b = n + 2 \), and \( c = n \):
\[ \frac{3}{5} = \frac{(n + 1)^2 + (n + 2)^2 - n^2}{2(n + 1)(n + 2)} \]

### Step 3: Expand and Simplify
Expanding the squares:
\[ (n + 1)^2 = n^2 + 2n + 1 \]
\[ (n + 2)^2 = n^2 + 4n + 4 \]
Thus,
\[ (n + 1)^2 + (n + 2)^2 - n^2 = (n^2 + 2n + 1) + (n^2 + 4n + 4) - n^2 = n^2 + 6n + 5 \]
Now substituting back:
\[ \frac{3}{5} = \frac{n^2 + 6n + 5}{2(n + 1)(n + 2)} \]

### Step 4: Cross Multiply
Cross multiplying gives:
\[ 3 \cdot 2(n + 1)(n + 2) = 5(n^2 + 6n + 5) \]
This simplifies to:
\[ 6(n^2 + 3n + 2) = 5(n^2 + 6n + 5) \]

### Step 5: Rearranging the Equation
Expanding both sides:
\[ 6n^2 + 18n + 12 = 5n^2 + 30n + 25 \]
Rearranging gives:
\[ 6n^2 - 5n^2 + 18n - 30n + 12 - 25 = 0 \]
\[ n^2 - 12n - 13 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-13)}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 + 52}}{2} = \frac{12 \pm \sqrt{196}}{2} = \frac{12 \pm 14}{2} \]
This gives:
\[ n = 13 \quad \text{or} \quad n = -1 \]
Since \( n \) must be positive, we take \( n = 13 \).

### Step 7: Find the Sides
Now substituting \( n \) back into the expressions for the sides:
\[ a = n + 1 = 14, \quad b = n + 2 = 15, \quad c = n = 13 \]

### Step 8: Calculate the Area
Using the formula for the area of a triangle:
\[ \Delta = \frac{1}{2}ab \sin C \]
We know \( \sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \).
Thus,
\[ \Delta = \frac{1}{2} \cdot 14 \cdot 15 \cdot \frac{4}{5} \]
Calculating this:
\[ \Delta = \frac{1}{2} \cdot 14 \cdot 15 \cdot \frac{4}{5} = \frac{1}{2} \cdot 14 \cdot 12 = 84 \]

### Final Answer
The area of triangle ABC is \( \Delta = 84 \).

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