Let's solve for the angle $x$ in the given triangle configuration.

Step 1: Analyze the Diagram

From the diagram:

• We have a triangle $ABC$ with point $D$ on side $BC$.
• The angles in the diagram are marked as follows:
  • $\angle BAC = 3x$
  • $\angle ABC = 2x$
  • $\angle ACB = x$
  • $\angle BDC = x$

Step 2: Use the Triangle Angle Sum Property

In any triangle, the sum of the interior angles is $180^\circ$. For $\triangle ABC$, we have: $\angle BAC + \angle ABC + \angle ACB = 180^\circ$ Substituting the given angles: $3x + 2x + x = 180^\circ$ Simplifying, we get: $6x = 180^\circ$ $x = 30^\circ$

Step 3: Verification of Angle $x$

To ensure accuracy, consider the angle $\angle BDC = x$, which also matches $30^\circ$ based on the geometric properties and angle relationships.

Conclusion:

The angle $x$ is $30^\circ$.

Would you like further details or have any questions?

Related Questions:

1. How does the angle $x = 30^\circ$ affect the other angles in the triangle?
2. Can the triangle $ABC$ be classified based on these angles?
3. What would happen if $\angle BAC$ was changed to $4x$?
4. How can the triangle's side lengths be determined using trigonometry if $x = 30^\circ$?
5. What is the role of point $D$ in the configuration of the triangle?
6. How does this solution utilize basic geometric properties?
7. Can this problem be solved using the Law of Sines or Cosines?
8. How would the problem change if $\angle ABC$ was $3x$ instead of $2x$?

Tip:

Always cross-check your work by summing the angles in each triangle to ensure they total $180^\circ$.