A central angle subtending the same arc is twice the corresponding inscribed angle.

Therefore,

\(\angle AOB=2\cdot 58^\circ=116^\circ\).

Answer: \(116^\circ\).

Since \(AB\) is a diameter, the arc \(AB\) of the upper semicircle measures \(180^\circ\).

Therefore, arc \(CB\) is

\(180^\circ-114^\circ=66^\circ\).

The inscribed angle \(\angle CAB\) subtends arc \(CB\), so it equals half of that arc:

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

Answer: \(33^\circ\).

The angle between two tangents equals \(180^\circ\) minus the central angle subtending the arc between the tangency points.

Therefore,

\(\angle AOB=180^\circ-38^\circ=142^\circ\).

Answer: \(142^\circ\).

The inscribed angle \(\angle CDB\) subtends arc \(CB\), so

\(\overset\frown{CB}=2\cdot 44^\circ=88^\circ\).

The inscribed angle \(\angle CAD\) subtends arc \(CD\), so

\(\overset\frown{CD}=2\cdot 42^\circ=84^\circ\).

Angle \(\angle BAD\) subtends arc \(BD\) not containing point \(A\). This arc consists of arcs \(BC\) and \(CD\):

\(\overset\frown{BD}=88^\circ+84^\circ=172^\circ\).

Hence

\(\angle BAD=\dfrac{172^\circ}{2}=86^\circ\).

Answer: \(86^\circ\).

An angle formed by two chords intersecting inside a circle equals half the sum of the arcs intercepted by this angle and its vertical angle.

Therefore,

\(\angle BND=\dfrac{\overset\frown{BD}+\overset\frown{AC}}{2}=\dfrac{88^\circ+74^\circ}{2}=81^\circ\).

Answer: \(81^\circ\).

The inscribed angle \(\angle DAB\) subtends arc \(DB\), so

\(\overset\frown{DB}=2\cdot 60^\circ=120^\circ\).

The inscribed angle \(\angle CDA\) subtends arc \(CA\), so

\(\overset\frown{CA}=2\cdot 20^\circ=40^\circ\).

Angle \(\angle ANB\) and its vertical angle subtend arcs \(AB\) and \(CD\), whose sum is

\(360^\circ-120^\circ-40^\circ=200^\circ\).

Therefore,

\(\angle ANB=\dfrac{200^\circ}{2}=100^\circ\).

Answer: \(100^\circ\).

The inscribed angle \(\angle NDA\) subtends arc \(NA\), so

\(\overset\frown{NA}=2\cdot 32^\circ=64^\circ\).

Since \(DA\) is a diameter, arc \(DNA\) measures \(180^\circ\).

Hence arc \(ND\) equals

\(180^\circ-64^\circ=116^\circ\).

Angle \(\angle NMD\) is an inscribed angle subtending arc \(ND\), so

\(\angle NMD=\dfrac{116^\circ}{2}=58^\circ\).

Answer: \(58^\circ\).

According to the marks in the diagram, arcs \(BC\) and \(CA\) are equal, so chords \(BC\) and \(CA\) are equal. Therefore triangle \(ABC\) is isosceles, and the base angles are equal:

\(\angle BAC=\angle ABC\).

The sum of the angles of a triangle is \(180^\circ\), hence

\(\angle BAC=\angle ABC=\dfrac{180^\circ-54^\circ}{2}=63^\circ\).

Answer: \(63^\circ\).

Since \(AB\) is a diameter, the arc of the upper semicircle measures \(180^\circ\).

Therefore, arc \(CB\) is

\(180^\circ-112^\circ=68^\circ\).

The inscribed angle \(\angle CAB\) subtends arc \(CB\), hence

\(\angle CAB=\dfrac{68^\circ}{2}=34^\circ\).

Answer: \(34^\circ\).

The angle between the tangents equals \(180^\circ\) minus the central angle subtending the arc between the points of tangency.

Therefore,

\(\angle AOB=180^\circ-113^\circ=67^\circ\).

Answer: \(67^\circ\).

Angle \(\angle CDB\) subtends arc \(CB\), so

\(\overset\frown{CB}=2\cdot 42^\circ=84^\circ\).

Angle \(\angle CAD\) subtends arc \(CD\), so

\(\overset\frown{CD}=2\cdot 45^\circ=90^\circ\).

Angle \(\angle BAD\) subtends arc \(BD\) not containing point \(A\):

\(\overset\frown{BD}=84^\circ+90^\circ=174^\circ\).

Therefore,

\(\angle BAD=\dfrac{174^\circ}{2}=87^\circ\).

Answer: \(87^\circ\).

An angle formed by two chords intersecting inside a circle equals half the sum of the corresponding arcs:

\(\angle BND=\dfrac{\overset\frown{AC}+\overset\frown{BD}}{2}=\dfrac{72^\circ+86^\circ}{2}=79^\circ\).

Answer: \(79^\circ\).

Angle \(\angle DAB\) subtends arc \(DB\), so

\(\overset\frown{DB}=2\cdot 42^\circ=84^\circ\).

Angle \(\angle CDA\) subtends arc \(CA\), so

\(\overset\frown{CA}=2\cdot 27^\circ=54^\circ\).

Then arcs \(AB\) and \(CD\) together measure

\(360^\circ-84^\circ-54^\circ=222^\circ\).

The angle between the intersecting chords is

\(\angle ANB=\dfrac{222^\circ}{2}=111^\circ\).

Answer: \(111^\circ\).

The inscribed angle \(\angle NDA\) subtends arc \(NA\), therefore

\(\overset\frown{NA}=2\cdot 36^\circ=72^\circ\).

Since \(DA\) is a diameter, arc \(DNA\) measures \(180^\circ\).

Then arc \(ND\) is

\(180^\circ-72^\circ=108^\circ\).

Angle \(\angle NMD\) is an inscribed angle subtending arc \(ND\), so

\(\angle NMD=\dfrac{108^\circ}{2}=54^\circ\).

Answer: \(54^\circ\).

The angle between two tangents equals \(180^\circ\) minus the central angle subtending the arc between the tangency points.

Therefore,

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

Answer: \(134^\circ\).

Angle \(\angle CDB\) subtends arc \(CB\), so

\(\overset\frown{CB}=2\cdot 46^\circ=92^\circ\).

Angle \(\angle CAD\) subtends arc \(CD\), so

\(\overset\frown{CD}=2\cdot 41^\circ=82^\circ\).

Angle \(\angle BAD\) subtends arc \(BD\) not containing point \(A\):

\(\overset\frown{BD}=92^\circ+82^\circ=174^\circ\).

Hence

\(\angle BAD=\dfrac{174^\circ}{2}=87^\circ\).

Answer: \(87^\circ\).