From the diagram, the left transversal makes a \(52^\circ\) angle with line \(a\), and the right transversal makes a \(68^\circ\) angle with line \(b\).

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

The angle between the transversals at the vertex is

\(x=180^\circ-52^\circ-68^\circ=60^\circ\).

Answer: \(60^\circ\).

From the diagram, the angle between the left transversal and line \(b\) is \(53^\circ\), and the angle between the transversals at the lower vertex is \(60^\circ\).

Let the acute angle of inclination of the right transversal to the parallel line be \(x\). Then

\(53^\circ+60^\circ+x=180^\circ\).

Therefore,

\(x=180^\circ-53^\circ-60^\circ=67^\circ\).

Answer: \(67^\circ\).

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

Let the smaller angle be \(x\), then the larger angle is \(x+46^\circ\).

We get:

\(x+(x+46^\circ)=180^\circ\).

\(2x=134^\circ\Rightarrow x=67^\circ\).

So the larger angle is

\(67^\circ+46^\circ=113^\circ\).

Answer: \(113^\circ\).

Adjacent angles of a parallelogram are supplementary.

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

So

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

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

Answer: \(36^\circ\).

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

Let the smaller angle be \(x\), then the larger one is \(x+58^\circ\).

We get

\(x+(x+58^\circ)=180^\circ\).

\(2x=122^\circ\Rightarrow x=61^\circ\).

Answer: \(61^\circ\).

Let the smaller angle be \(x\), then the larger angle is \(5x\).

Adjacent angles along a leg of the trapezoid are supplementary:

\(x+5x=180^\circ\).

\(6x=180^\circ\Rightarrow x=30^\circ\).

So the larger angle is

\(5\cdot 30^\circ=150^\circ\).

Answer: \(150^\circ\).

In a rhombus, adjacent angles are supplementary, so

\(\angle BCD=180^\circ-134^\circ=46^\circ\).

The diagonal \(AC\) of a rhombus bisects the angle at vertex \(C\).

Therefore,

\(x=\dfrac{46^\circ}{2}=23^\circ\).

Answer: \(23^\circ\).

The diagonal \(AC\) of a rhombus bisects the angle at vertex \(A\).

So

\(\angle DAB=2\cdot 28^\circ=56^\circ\).

Adjacent angles of a rhombus are supplementary, therefore

\(x=180^\circ-56^\circ=124^\circ\).

Answer: \(124^\circ\).

In triangle \(ABC\), angle \(C\) is a right angle, i.e. \(90^\circ\), and angle \(A=32^\circ\).

Hence

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

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

\(x=\dfrac{58^\circ}{2}=29^\circ\).

Answer: \(29^\circ\).

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

The vertex angle at \(C\) is \(46^\circ\).

Therefore each base angle equals

\(\dfrac{180^\circ-46^\circ}{2}=67^\circ\).

Since \(BD\) is the bisector of angle \(ABC\), it halves this angle:

\(x=\dfrac{67^\circ}{2}=33.5^\circ\).

Answer: \(33.5^\circ\).

From the diagram, angle \(C\) is \(52^\circ\), and the angle between the bisector \(BD\) and side \(BC\) is \(25^\circ\).

Since \(BD\) is a bisector, the whole angle \(B\) is

\(2\cdot 25^\circ=50^\circ\).

Then

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

Answer: \(78^\circ\).

From the diagram, one half of angle \(ABC\) equals \(31^\circ\).

Since \(BD\) is a bisector, the whole angle \(B\) is

\(2\cdot 31^\circ=62^\circ\).

Angle \(C\) is a right angle, i.e. \(90^\circ\). Hence

\(x=180^\circ-90^\circ-62^\circ=28^\circ\).

Answer: \(28^\circ\).

In the diagram, angle \(C=66^\circ\), and one half of angle \(B\) is \(27^\circ\).

Since \(BD\) is a bisector, in triangle \(CBD\) the angle at vertex \(B\) equals \(27^\circ\).

Therefore the required angle at vertex \(D\) is

\(x=180^\circ-66^\circ-27^\circ=87^\circ\).

Answer: \(87^\circ\).

Diagonal \(AC\) divides the parallelogram into triangle \(ABC\).

In triangle \(ABC\), the known angles are:

\(\angle BAC=33^\circ\), \(\angle BCA=17^\circ\).

Then

\(\angle ABC=180^\circ-33^\circ-17^\circ=130^\circ\).

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 and adjacent angles are supplementary.

From the diagram, \(\angle D=127^\circ\), hence \(\angle B=127^\circ\).

In triangle \(ABC\), formed by diagonal \(AC\), we know:

\(\angle A=32^\circ\), \(\angle B=127^\circ\).

Then

\(x=180^\circ-32^\circ-127^\circ=21^\circ\).

Answer: \(21^\circ\).

From the diagram, the angle between bisector \(BD\) and side \(BC\) equals \(27^\circ\).

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

\(2\cdot 27^\circ=54^\circ\).

Altitude \(AN\) is perpendicular to side \(BC\), so \(\angle ANB=90^\circ\).

In triangle \(ABN\):

\(x+54^\circ+90^\circ=180^\circ\).

Therefore,

\(x=36^\circ\).

Answer: \(36^\circ\).