\(15^{2}=6^{2}+x^{2}-2(6)(x)\cos (30^{\circ })\)

We are given the equation:

\[
15^{2} = 6^{2} + x^{2} - 2(6)(x) \cos(30^\circ)
\]

### Step 1: Simplify both sides

We start by calculating the known values:

\[
15^2 = 225, \quad 6^2 = 36, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}
\]

So, the equation becomes:

\[
225 = 36 + x^2 - 2(6)(x) \cdot \frac{\sqrt{3}}{2}
\]

### Step 2: Simplify further

The \( 2 \)'s cancel out in the third term:

\[
225 = 36 + x^2 - 6x\sqrt{3}
\]

### Step 3: Rearrange the equation

Move the constant 36 to the left side:

\[
225 - 36 = x^2 - 6x\sqrt{3}
\]

\[
189 = x^2 - 6x\sqrt{3}
\]

### Step 4: Solve the quadratic equation

This is a quadratic equation in the form:

\[
x^2 - 6x\sqrt{3} - 189 = 0
\]

We can solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( x^2 - 6x\sqrt{3} - 189 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = -6\sqrt{3} \)
- \( c = -189 \)

Now let's plug these values into the quadratic formula:

\[
x = \frac{-(-6\sqrt{3}) \pm \sqrt{(-6\sqrt{3})^2 - 4(1)(-189)}}{2(1)}
\]

Let me calculate that for you.The solutions to the quadratic equation are approximately:

\[
x_1 \approx 19.89 \quad \text{and} \quad x_2 \approx -9.50
\]

Since distances are usually positive, the most likely solution is:

\[
x \approx 19.89
\]

Would you like any further details or explanations?

Here are five related questions for further exploration:

1. How is the cosine law used in solving triangle-related problems?
2. Can the quadratic formula solve all quadratic equations?
3. What is the significance of the discriminant in the quadratic formula?
4. How does the angle in a cosine term affect the length of a triangle's sides?
5. What other methods can be used to solve quadratic equations?

**Tip:** Always check the discriminant when using the quadratic formula to know if you have real or complex solutions.

15^{2}=6^{2}+x^{2}-2(6)(x)\cos (30^{\circ })

Learn how to solve the equation 15^2 = 6^2 + x^2 - 2(6)(x) cos(30°) using the cosine law and quadratic formula. This problem involves key concepts from trigonometry and algebra, suitable for high school students in Grades 10-12.

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