To solve for $x$ in this triangle problem, let's analyze step by step.

1. Use the formula for the area of a triangle:

The general formula for the area of a triangle with two sides $a$ and $b$ and included angle $\theta$ is: $\text{Area} = \frac{1}{2} \times a \times b \times \sin(\theta)$

Here:

• One side $a = x$,
• The other side $b = x + 4$,
• The included angle $\theta = 60^\circ$,
• The area is $15\sqrt{3} \, \text{cm}^2$.

2. Substitute into the area formula:

$15\sqrt{3} = \frac{1}{2} \times x \times (x + 4) \times \sin(60^\circ)$ We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so substituting: $15\sqrt{3} = \frac{1}{2} \times x \times (x + 4) \times \frac{\sqrt{3}}{2}$

3. Simplify the equation:

Combine terms: $15\sqrt{3} = \frac{\sqrt{3}}{4} \times x \times (x + 4)$

To eliminate $\sqrt{3}$ from both sides, divide through by $\sqrt{3}$: $15 = \frac{1}{4} \times x \times (x + 4)$

4. Clear the fraction:

Multiply through by 4 to simplify: $60 = x(x + 4)$

Expand the right-hand side: $60 = x^2 + 4x$

5. Form a quadratic equation:

Rearrange to standard quadratic form: $x^2 + 4x - 60 = 0$

6. Solve the quadratic equation:

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = 1$,
• $b = 4$,
• $c = -60$.

Substitute into the formula: $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-60)}}{2(1)}$

Simplify step by step: $x = \frac{-4 \pm \sqrt{16 + 240}}{2}$ $x = \frac{-4 \pm \sqrt{256}}{2}$ $x = \frac{-4 \pm 16}{2}$

7. Calculate the two solutions:

• For $-4 + 16$: $x = \frac{12}{2} = 6$
• For $-4 - 16$: $x = \frac{-20}{2} = -10$

Since $x$ represents a side length, it must be positive: $x = 6$

Final Answer:

The value of $x$ is 6 cm.

5 Related Questions:

1. How do you calculate the area of a triangle using trigonometry and side lengths?
2. What is the importance of the sine rule in triangles?
3. How do you solve a quadratic equation step by step?
4. What is the relationship between angles and side lengths in non-right triangles?
5. How do you calculate the sine of angles like $30^\circ, 45^\circ, 60^\circ$?

Tip:

When working with trigonometry in triangles, always check if the triangle involves special angles like $30^\circ, 45^\circ,$ or $60^\circ$, as these simplify calculations.