The left transversal forms a \(55^\circ\) angle with line \(a\), and the right transversal forms a \(63^\circ\) angle with line \(b\).

Since lines \(a\) and \(b\) are parallel, these are the acute inclination angles of the transversals to the parallel lines.

The angle at the vertex between the transversals is:

\(x=180^\circ-55^\circ-63^\circ=62^\circ\).

Answer: \(62^\circ\).

One transversal forms a \(58^\circ\) angle with the parallel line, and the other forms a \(62^\circ\) angle.

Therefore, the angle between the transversals at the vertex is:

\(x=180^\circ-58^\circ-62^\circ=60^\circ\).

Answer: \(60^\circ\).

Adjacent angles of a parallelogram sum to \(180^\circ\).

Let the smaller angle be \(x\), then the larger one is \(x+30^\circ\).

Set up the equation:

\(x+(x+30^\circ)=180^\circ\).

\(2x=150^\circ\Rightarrow x=75^\circ\).

So the larger angle is:

\(75^\circ+30^\circ=105^\circ\).

Answer: \(105^\circ\).

Adjacent angles of a parallelogram sum to \(180^\circ\).

Let the smaller angle be \(x\), then the larger angle is \(3x\).

So

\(x+3x=180^\circ\).

\(4x=180^\circ\Rightarrow x=45^\circ\).

Answer: \(45^\circ\).

In an isosceles trapezoid, the angles at the same base are equal, and adjacent angles along a leg sum to \(180^\circ\).

Let the smaller angle be \(x\), then the larger one is \(x+40^\circ\).

Then

\(x+(x+40^\circ)=180^\circ\).

\(2x=140^\circ\Rightarrow x=70^\circ\).

Answer: \(70^\circ\).

Let the smaller angle be \(x\), then the larger one is \(4x\).

Adjacent angles along a leg are supplementary:

\(x+4x=180^\circ\).

\(5x=180^\circ\Rightarrow x=36^\circ\).

Hence the larger angle is

\(4\cdot 36^\circ=144^\circ\).

Answer: \(144^\circ\).

In a rhombus, opposite angles are equal and adjacent angles are supplementary.

If \(\angle B=130^\circ\), then

\(\angle C=180^\circ-130^\circ=50^\circ\).

A diagonal of a rhombus bisects the angles at its endpoints. Therefore, diagonal \(AC\) bisects angle \(C\).

Hence

\(x=\dfrac{50^\circ}{2}=25^\circ\).

Answer: \(25^\circ\).

In a rhombus, diagonal \(AC\) bisects the angles at vertices \(A\) and \(C\).

Therefore, if \(\angle CAD=24^\circ\), then the whole angle at vertex \(A\) is

\(\angle DAB=2\cdot 24^\circ=48^\circ\).

Adjacent angles of a rhombus are supplementary, so

\(x=\angle ABC=180^\circ-48^\circ=132^\circ\).

Answer: \(132^\circ\).

In right triangle \(ABC\), angle \(C\) equals \(90^\circ\), and angle \(A\) equals \(34^\circ\).

So angle \(B\) is

\(\angle B=180^\circ-90^\circ-34^\circ=56^\circ\).

Since \(BD\) is the bisector of angle \(B\), it divides this angle into two equal parts:

\(x=\dfrac{56^\circ}{2}=28^\circ\).

Answer: \(28^\circ\).

Since triangle \(ABC\) is isosceles with base \(AB\), the base angles are equal.

From the diagram, the vertex angle at \(C\) is \(48^\circ\).

Therefore each base angle is

\(\dfrac{180^\circ-48^\circ}{2}=66^\circ\).

Since \(BD\) is the bisector of angle \(ABC\), we have

\(x=\dfrac{66^\circ}{2}=33^\circ\).

Answer: \(33^\circ\).

From the diagram, the angle at vertex \(C\) is \(50^\circ\), and half of the angle at vertex \(B\) is \(26^\circ\), since \(BD\) is a bisector.

Therefore,

\(\angle B=2\cdot 26^\circ=52^\circ\).

Then angle \(A\) is

\(x=180^\circ-50^\circ-52^\circ=78^\circ\).

Answer: \(78^\circ\).

From the diagram, one half of angle \(ABC\) is \(29^\circ\). Since \(BD\) is a bisector, the whole angle \(B\) equals

\(58^\circ\).

Triangle \(ABC\) is right-angled, so

\(x=\angle A=180^\circ-90^\circ-58^\circ=32^\circ\).

Answer: \(32^\circ\).

From the diagram, the angle at vertex \(C\) is \(64^\circ\), and bisector \(BD\) divides angle \(B\) into two equal parts, one of which is \(28^\circ\).

Therefore, in triangle \(CBD\):

\(\angle CBD=28^\circ,\quad \angle BCD=64^\circ\).

Then

\(x=\angle CDB=180^\circ-64^\circ-28^\circ=88^\circ\).

Answer: \(88^\circ\).

Consider triangle \(ABC\) formed by diagonal \(AC\).

From the diagram,

\(\angle BAC=32^\circ,\quad \angle BCA=18^\circ\).

Then

\(\angle ABC=180^\circ-32^\circ-18^\circ=130^\circ\).

In a parallelogram, opposite angles are equal, so

\(x=\angle D=\angle B=130^\circ\).

Answer: \(130^\circ\).

In a parallelogram, opposite angles are equal, so

\(\angle B=\angle D=123^\circ\).

Diagonal \(AC\) forms an angle of \(34^\circ\) at vertex \(A\) in triangle \(ABC\), and the required angle \(x\) is at vertex \(C\).

Thus, in triangle \(ABC\):

\(34^\circ+123^\circ+x=180^\circ\).

Hence

\(x=23^\circ\).

Answer: \(23^\circ\).

From the diagram, the angle between bisector \(BD\) and side \(BC\) is \(26^\circ\).

Since \(BD\) is the bisector of angle \(B\), we get

\(\angle B=2\cdot 26^\circ=52^\circ\).

Altitude \(AN\) is perpendicular to side \(BC\), so the angle between \(AN\) and \(AB\) is complementary to angle \(B\):

\(x=90^\circ-52^\circ=38^\circ\).

Answer: \(38^\circ\).