To solve for $x$, we will break the problem down step by step, focusing on the relationships within the two smaller right triangles, $\triangle BDC$ and $\triangle ADC$, created by the perpendicular line $CD$.

Step 1: Understanding the Given Information

We are given a right triangle $ABC$, where:

• $\angle ACB = 90^\circ$
• $\angle CBD = 30^\circ$
• $\angle CAD = 60^\circ$
• $CB = 2\sqrt{3}$ units
• $AD = x$ units
• The line segment $CD$ is perpendicular to $AB$

From this setup, we know that triangle $ABC$ is a right triangle, and $CD$ creates two smaller right triangles.

Step 2: Using Trigonometry in $\triangle BDC$

Since $\angle CBD = 30^\circ$ and $\triangle BDC$ is a right triangle, we can apply trigonometry to relate the sides.

In right triangle $BDC$: $\tan(\angle CBD) = \frac{BC}{BD}$ Since $\angle CBD = 30^\circ$, and $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we get: $\frac{2\sqrt{3}}{BD} = \frac{1}{\sqrt{3}}$ Now, solve for $BD$: $BD = 2\sqrt{3} \times \sqrt{3} = 6$

Step 3: Using Trigonometry in $\triangle ADC$

In $\triangle ADC$, we know $\angle CAD = 60^\circ$, and $\triangle ADC$ is a right triangle. We can use the fact that: $\tan(\angle CAD) = \frac{AC}{AD}$ Since $\tan(60^\circ) = \sqrt{3}$, we get: $\frac{AC}{x} = \sqrt{3}$ Thus, $AC = x \sqrt{3}$

Step 4: Applying the Pythagorean Theorem in $\triangle ABC$

Now, let's apply the Pythagorean theorem in $\triangle ABC$ to relate $AB$, $AC$, and $BC$. Since $\angle ACB = 90^\circ$, we know: $AB^2 = AC^2 + BC^2$ We already know $BC = 2\sqrt{3}$, so: $AB^2 = AC^2 + (2\sqrt{3})^2$ $AB^2 = AC^2 + 12$

Next, we can express $AB$ as the sum of $AD$ and $BD$: $AB = AD + BD = x + 6$ Substitute this into the Pythagorean theorem: $(x + 6)^2 = AC^2 + 12$

Step 5: Substituting $AC = x \sqrt{3}$

Now, substitute $AC = x\sqrt{3}$ into the equation: $(x + 6)^2 = (x\sqrt{3})^2 + 12$ $(x + 6)^2 = 3x^2 + 12$ Expanding the left side: $x^2 + 12x + 36 = 3x^2 + 12$ Now, move all terms to one side: $x^2 + 12x + 36 - 3x^2 - 12 = 0$ $-2x^2 + 12x + 24 = 0$ Divide through by -2: $x^2 - 6x - 12 = 0$

Step 6: Solving the Quadratic Equation

Now, solve the quadratic equation $x^2 - 6x - 12 = 0$ using the quadratic formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-12)}}{2(1)}$ $x = \frac{6 \pm \sqrt{36 + 48}}{2}$ $x = \frac{6 \pm \sqrt{84}}{2}$ $x = \frac{6 \pm 2\sqrt{21}}{2}$ $x = 3 \pm \sqrt{21}$

Since $x$ represents a length, it must be positive. Thus, the solution is: $x = 3 + \sqrt{21}$

Final Answer:

The value of $x$ is $3 + \sqrt{21}$ units.

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Would you like further details or clarification? Here are some related questions:

1. How does the Pythagorean theorem apply to right triangles in geometry?
2. Can trigonometric ratios be used in any triangle, or just right triangles?
3. What are other ways to solve quadratic equations besides the quadratic formula?
4. How does the tangent function relate to the sides of a right triangle?
5. How can geometric figures be decomposed into smaller triangles for easier analysis?

Tip: Remember that when solving geometric problems with trigonometry or the Pythagorean theorem, organizing your steps systematically and using known ratios and relationships can simplify the process.