Question

what is the value of ( x ) in the figure?

Explanation:

Step1: Prove triangle congruence

In \(\triangle ABC\) and \(\triangle BDC\), \(AC = BD = 4\), \(AB = CD = 3.9\), \(BC=CB\) (common side). By SSS (Side - Side - Side) congruence criterion, \(\triangle ABC\cong\triangle BDC\).

Step2: Use triangle angle - sum property

In \(\triangle ABC\), by the angle - sum property of a triangle (\(\angle A+\angle ABC+\angle BCA = 180^{\circ}\)). Given \(\angle A = 100^{\circ}\), \(\angle BCA=30^{\circ}\), then \(\angle ABC=180^{\circ}-(100^{\circ} + 30^{\circ})=50^{\circ}\).
Since \(\triangle ABC\cong\triangle BDC\), \(\angle CBD=\angle BCA = 30^{\circ}\), and \(\angle ABD=\angle ABC\). Also, in \(\triangle BCD\), \(\angle BDC=\angle A = 100^{\circ}\). Using the angle - sum property in \(\triangle BCD\) (\(\angle BDC+\angle DBC+\angle BCD=180^{\circ}\)), and we know \(\angle BCD=x + 30^{\circ}\).
Another way: Since \(\triangle ABC\cong\triangle BDC\), \(\angle DBC=\angle BCA = 30^{\circ}\). In \(\triangle ABC\), \(\angle ABC = 50^{\circ}\). And because of the congruence, in the quadrilateral (or using the congruent triangles relation directly), we can also use the property that alternate interior angles (if we consider the parallel - like property from congruent sides). But more simply, since \(\triangle ABC\cong\triangle BDC\), and using the fact that in \(\triangle ABC\) we found \(\angle ABC = 50^{\circ}\) (from \(180-(100 + 30)\)), and in \(\triangle BCD\), assume the triangle properties. Wait, a better approach:
Since \(AB = CD\), \(AC = BD\), \(BC = CB\), \(\triangle ABC\cong\triangle DCB\) (SSS). Then \(\angle ACB=\angle DBC = 30^{\circ}\), \(\angle BAC=\angle CDB=100^{\circ}\). In \(\triangle ABC\), \(\angle ABC=180-(100 + 30)=50^{\circ}\). And since \(\triangle ABC\cong\triangle DCB\), \(\angle ABC=\angle DCB\). But \(\angle DCB=x + 30^{\circ}\). Also, in \(\triangle BCD\), \(\angle CBD = 30^{\circ}\), \(\angle CDB = 100^{\circ}\), so \(\angle BCD=180-(100 + 30)=50^{\circ}\). Then \(x+30^{\circ}=50^{\circ}\).

Step3: Solve for \(x\)

Subtract \(30^{\circ}\) from both sides of the equation \(x + 30^{\circ}=50^{\circ}\). We get \(x=50^{\circ}-30^{\circ}=20^{\circ}\).

Answer:

\(20\)