In a circle centered at $O$, a diameter $AB$ is extended to a point $C$ outside the circle. If $CD = 7\sqrt{3} ext{ cm}$ and $\angle DBC = 120^\circ$, then what is the length of the diameter of the circle?

Question

In a circle centered at $O$, a diameter $AB$ is extended to a point $C$ outside the circle. If $CD = 7\sqrt{3} 	ext{ cm}$ and $\angle DBC = 120^\circ$, then what is the length of the diameter of the circle?

Accepted Answer

**Solution:**

1. Let the radius of the circle be $r$. Since $AB$ is the diameter, $AB = 2r$.

2. Given $CD = 7\sqrt{3} 	ext{ cm}$ and $\angle DBC = 120^\circ$.

3. Since $CD$ is tangent to the circle at $D$, the radius $OD$ is perpendicular to $CD$, so $\angle ODC = 90^\circ$.

4. Consider triangle $ODB$ where $OB = OD = r$ (radii of the circle).

5. The angle $\angle OBD$ can be found using the straight line property:
   $$\angle OBD = 180^\circ - \angle DBC = 180^\circ - 120^\circ = 60^\circ$$

6. Since $OB = OD$, triangle $ODB$ is isosceles with $\angle OBD = \angle ODB = 60^\circ$.

7. Therefore, triangle $ODB$ is equilateral, and all sides are equal to $r$.

8. In triangle $ODC$, angles are:
   - $\angle ODC = 90^\circ$ (radius perpendicular to tangent)
   - $\angle ODB = 60^\circ$ (from above)
   - So, $\angle OCD = 30^\circ$ (since sum of angles in triangle is $180^\circ$)

9. Using right triangle $ODC$, apply the tangent function:
   $$	an(\angle OCD) = \frac{OD}{CD} = \frac{r}{7\sqrt{3}}$$

10. Since $	an 30^\circ = \frac{1}{\sqrt{3}}$, we have:
    $$\frac{1}{\sqrt{3}} = \frac{r}{7\sqrt{3}}$$

11. Solving for $r$:
    $$r = 7 	ext{ cm}$$

12. Diameter $AB = 2r = 2 	imes 7 = 14 	ext{ cm}$.

**Answer:** $14 	ext{ cm}$

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