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The problem in the image involves an isosceles triangle \(ABC\) where \( \overline{AB} = \overline{AC} = 4 \), and the area of the triangle is given as 12. The goal is to determine the value of \( \cos(\alpha) \), where \( \alpha \) is the angle \( \angle BAC \).

### Step 1: Finding the height of the triangle
The area of a triangle can be expressed as:

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

In this isosceles triangle, let \(h\) represent the height drawn from vertex \(A\) perpendicular to base \( \overline{BC} \). Since the triangle is isosceles, the height divides \( \overline{BC} \) into two equal segments, so each segment has length \( \frac{\overline{BC}}{2} \).

The area is given as 12, and the base \(BC\) is denoted by \(b\). The area formula gives us:

\[
12 = \frac{1}{2} \times b \times h
\]

This simplifies to:

\[
b \times h = 24 \tag{1}
\]

### Step 2: Using the Pythagorean theorem
Since the triangle is isosceles, we can use the Pythagorean theorem to express \(h\) in terms of \(b\). The relationship for half of the base (i.e., \( \frac{b}{2} \)) is:

\[
\left( \frac{b}{2} \right)^2 + h^2 = 4^2 = 16
\]

Simplifying:

\[
\frac{b^2}{4} + h^2 = 16 \tag{2}
\]

### Step 3: Solving for \(b\) and \(h\)
Now, we have two equations: (1) \( b \times h = 24 \) and (2) \( \frac{b^2}{4} + h^2 = 16 \). Solving this system will give us the values of \(b\) and \(h\).

Let me solve these equations for you.It seems the initial approach using symbolic computation gave overly complex results. Let's simplify the system manually.

We already have two equations:
1. \( b \times h = 24 \)
2. \( \frac{b^2}{4} + h^2 = 16 \)

We can express \(h\) from the first equation as:

\[
h = \frac{24}{b}
\]

Substitute this into the second equation:

\[
\frac{b^2}{4} + \left( \frac{24}{b} \right)^2 = 16
\]

Let me now simplify this equation further and solve for \(b\).It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can manually solve this equation:

\[
\frac{b^2}{4} + \left( \frac{24}{b} \right)^2 = 16
\]

Once you find \(b\), substitute it back into \( h = \frac{24}{b} \) to get the height. Afterward, use trigonometry to find \( \cos(\alpha) \).

If you'd like me to walk through the remaining steps in more detail, feel free to ask!

In the figure, an isosceles triangle [ABC] is shown. It is given that AB = 4, AC = BC, and the area of triangle [ABC] is 12. The amplitude of angle BAC is represented by α. Find cos(α).

Learn how to calculate the cosine of angle α in an isosceles triangle where the sides and area are provided. This problem utilizes concepts from trigonometry and the Pythagorean theorem. Ideal for high school students studying geometry and trigonometry.

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