An isosceles triangle has a vertex angle of 30 degrees, and each of the two equal sides measures 15 units. Find the length of the base of the triangle using trigonometric functions. Then, calculate the area of the triangle. Finally, suppose a circumcircle passes through all three vertices of the triangle. Find the radius of this circumcircle.

To solve the problem involving the isosceles triangle, we will go step by step:

### 1. **Length of the Base of the Triangle**

The isosceles triangle has two equal sides of 15 units each, and the vertex angle is 30 degrees. To find the length of the base, we can split the triangle into two right triangles by drawing an altitude from the vertex to the midpoint of the base. This altitude will bisect the base and the vertex angle, creating two right triangles with a 15-unit hypotenuse and a 15-degree angle (since the vertex angle is halved).

Using trigonometry, specifically the sine function, we can find half the length of the base:

\[
\sin(15^\circ) = \frac{\text{half of the base}}{15}
\]

Solving for half of the base:

\[
\text{half of the base} = 15 \cdot \sin(15^\circ) \approx 15 \cdot 0.2588 \approx 3.882
\]

Thus, the full base is:

\[
\text{base} = 2 \cdot 3.882 \approx 7.764 \text{ units}
\]

### 2. **Area of the Triangle**

To find the area of the triangle, we use the formula for the area of a triangle:

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

We already know the base is approximately 7.764 units. Now, we need to calculate the height, which is the opposite side of the right triangle we created. Using the cosine function:

\[
\cos(15^\circ) = \frac{\text{height}}{15}
\]

Solving for the height:

\[
\text{height} = 15 \cdot \cos(15^\circ) \approx 15 \cdot 0.9659 \approx 14.488 \text{ units}
\]

Now, we can calculate the area:

\[
\text{Area} = \frac{1}{2} \cdot 7.764 \cdot 14.488 \approx 56.22 \text{ square units}
\]

### 3. **Radius of the Circumcircle**

The circumradius \( R \) of a triangle can be found using the formula:

\[
R = \frac{abc}{4A}
\]

where \( a \), \( b \), and \( c \) are the sides of the triangle, and \( A \) is the area of the triangle. Since the triangle is isosceles, two of the sides are equal to 15, and the base \( c \) is 7.764 units. The area \( A \) is approximately 56.22 square units.

Substitute these values into the formula:

\[
R = \frac{15 \cdot 15 \cdot 7.764}{4 \cdot 56.22} \approx \frac{1742.94}{224.88} \approx 7.75 \text{ units}
\]

Thus, the radius of the circumcircle is approximately \( 7.75 \) units.

### Summary of Results:
1. **Base length**: \( \approx 7.764 \) units
2. **Area**: \( \approx 56.22 \) square units
3. **Circumradius**: \( \approx 7.75 \) units

Would you like more details on any of these steps?

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Here are five related questions to deepen understanding:
1. How would the base length change if the vertex angle were increased?
2. What if the equal sides were shortened to 10 units—how would the area and circumradius change?
3. Can the height be determined using only the Pythagorean theorem instead of trigonometry?
4. How does the circumradius formula relate to the type of triangle (equilateral, scalene, etc.)?
5. If you were to inscribe a circle within the triangle, what would the radius of that circle be?

**Tip**: Understanding the relationship between a triangle's angles and its side lengths through trigonometric functions is a powerful tool for solving many geometric problems.

Solve an isosceles triangle with a 30-degree vertex angle and sides of 15 units using trigonometric functions. Find the base, calculate the area, and determine the circumradius of the triangle. Step-by-step solution provided, suitable for Grades 10-12.

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