Remember that the angle for a sector or arc is found as a percentage of the total $\displaystyle 360$ degrees of the circle.  The proportion of $\displaystyle x$ to $\displaystyle 360$ is the same as $\displaystyle 9.31$ to the total circumference of the circle.

The circumference of a circle is found by:

$\displaystyle C = 2*\pi*r$

For our data, this means:

$\displaystyle C = 2*9.1*\pi=18.2\pi$

Now we can solve for $\displaystyle x$ using the proportions:

$\displaystyle \frac{x}{360} = \frac{9.31}{18.2\pi}$

Cross multiply:

$\displaystyle 18.2x\pi=3351.6$

Divide both sides by $\displaystyle 18.2\pi$:

$\displaystyle x=58.61798980953807$

Therefore, $\displaystyle x$ is $\displaystyle 58.62$˚.

Remember that the angle for a sector or arc is found as a percentage of the total $\displaystyle 360$ degrees of the circle.  The proportion of $\displaystyle x$ to $\displaystyle 360$ is the same as $\displaystyle 4.5\pi$ to the total area of the circle.

The area of a circle is found by:

$\displaystyle A = \pi * r^2$

For our data, this means:

$\displaystyle A = 7.5^2\pi = 56.25\pi$

Now we can solve for $\displaystyle x$ using the proportions:

$\displaystyle \frac{x}{360} = \frac{2.81\pi}{56.25\pi}$

Cross multiply:

$\displaystyle 56.25x\pi=1011.6\pi$

Divide both sides by $\displaystyle 56.25\pi$:

$\displaystyle x=17.984$

Therefore, $\displaystyle x$ is $\displaystyle 17.98$˚.

Remember that the angle for a sector or arc is found as a percentage of the total $\displaystyle 360$ degrees of the circle.  The proportion of $\displaystyle x$ to $\displaystyle 360$ is the same as $\displaystyle 4.5\pi$ to the total area of the circle.

The area of a circle is found by:

$\displaystyle A = \pi * r^2$

For our data, this means:

$\displaystyle A = 6^2\pi = 36\pi$

Now we can solve for $\displaystyle x$ using the proportions:

$\displaystyle \frac{x}{360} = \frac{4.5\pi}{36\pi}$

Cross multiply:

$\displaystyle 36x\pi=1620\pi$

Divide both sides by $\displaystyle 36\pi$:

$\displaystyle x=45$

Therefore, $\displaystyle x$ is $\displaystyle 45$˚.

$\displaystyle \bigtriangleup ABC$ has angles of degree measure 30 and 60; the third angle must measure 90 degrees, making $\displaystyle \bigtriangleup ABC$ a right triangle.

For the sake of simplicity, let $\displaystyle BC = 1$; the reasoning is independent of the actual length. The smaller leg of a 30-60-90 triangle has length equal to $\displaystyle \frac{\sqrt{3}}{3}$ times that of the longer leg; this is about

$\displaystyle AB = \frac{\sqrt{3}}{3} \approx \frac{1.7}{3} \approx 0.57$

 The area of a right triangle is half the product of its legs, so 

$\displaystyle A = \frac{1}{2} \cdot 1 \cdot 0.57 = 0.5 \cdot 0.57 \approx 0.285$

Also, if $\displaystyle BC = 1$, then the orange semicircle has diameter 1 and radius $\displaystyle \frac{1}{2}$. Its area can be found by substituting $\displaystyle r = \frac{1}{2}$ in the formula:

$\displaystyle A = \frac{1}{2} \cdot \pi r^{2}$

$\displaystyle = \frac{1}{2} \cdot \pi \left ( \frac{1}{2} \right ) ^{2}$

$\displaystyle = \frac{1}{2} \cdot \pi \cdot \left ( \frac{1}{4 } \right )$

$\displaystyle = \frac{1}{8} \cdot \pi$

$\displaystyle \approx 0. 125 \cdot 3.14$

$\displaystyle \approx 0. 3925$

The orange semicircle has a greater area than $\displaystyle \bigtriangleup ABC$

That $\displaystyle \overline{AD}$ is a diameter of the circle is actually irrelevant to the problem. Two inscribed angles of a circle that both intercept the same arc, as $\displaystyle \angle CAD$ and $\displaystyle \angle CBD$ both do here, have the same measure.

The figure referenced is below:

$\displaystyle \overarc {ABC }$ is a semicircle, so $\displaystyle \overarc {ADC }$ is one as well; as a semicircle, its measure is $\displaystyle 180 ^{\circ }$. The inscribed angle that intercepts this semicircle, $\displaystyle \angle B$, is a right angle, of measure $\displaystyle 90 ^{\circ }$. $\displaystyle m \angle A = 40 ^{\circ }$, and the sum of the measures of the interior angles of a triangle is $\displaystyle 180 ^{\circ }$, so 

$\displaystyle m \angle C = 180 ^{\circ } - (m \angle A + m \angle B )$

$\displaystyle = 180 ^{\circ } - (40 ^{\circ } +90 ^{\circ } )$

$\displaystyle = 180 ^{\circ } - 130 ^{\circ }$

$\displaystyle = 50 ^{\circ }$

$\displaystyle \angle C$ has greater measure than $\displaystyle \angle A$, so the minor arc intercepted by  $\displaystyle \angle C$, which is $\displaystyle \overarc {BA}$, has greater measure than that intercepted by $\displaystyle \angle A$, which is $\displaystyle \overarc {BC}$. It follows that the major arc corresponding to the latter, which is $\displaystyle \overarc {BA C}$, has greater measure than that  corresponding to the former, which is $\displaystyle \overarc {BC A }$.

Construct $\displaystyle \overline{OD}$. The new figure is below:

$\displaystyle m \overarc {AD} = 60 ^{\circ }$, so $\displaystyle m \overarc {AC} < 60 ^{\circ }$. It follows that their respective central angles have measures

$\displaystyle m \angle AOD = 60 ^{\circ }$

and

$\displaystyle m \angle AOC < 60 ^{\circ }$.

Also, since $\displaystyle m \overarc {AD} = 60 ^{\circ }$ and $\displaystyle m \overarc {ACB} = 180 ^{\circ }$ - $\displaystyle \overarc {ACB}$ being a semicircle - by the Arc Addition Principle, $\displaystyle m \overarc {BD} = 120^{\circ }$. $\displaystyle \angle DAO$, an inscribed angle which intercepts this arc, has half this measure, which is $\displaystyle 60 ^{\circ }$. The other angle of $\displaystyle \bigtriangleup ADO$, which is $\displaystyle \angle ADO$, also measures $\displaystyle 60 ^{\circ }$, so  $\displaystyle \bigtriangleup ADO$ is equilateral.

$\displaystyle OD = OC$, since all radii are congruent;

$\displaystyle OA = OA$ by reflexivity;

$\displaystyle m \angle AOC < m \angle AOD$

By the Side-Angle-Side Inequality Theorem (or Hinge Theorem), it follows that $\displaystyle AC < AD$. Since $\displaystyle \bigtriangleup ADO$ is equilateral, $\displaystyle AD = OD$, and since all radii are congruent, $\displaystyle OD = OC$. Substituting, it follows that $\displaystyle AC < OC$.

Below is the inscribed trapezoid referenced, along with its diagonals.

$\displaystyle \overline{AD} ||\overline{BC}$, so, by the Alternate Interior Angles Theorem, 

$\displaystyle \angle ACB \cong \angle CAD$, and their intercepted angles are also congruent - that is,

$\displaystyle m \overarc{AB} = m \overarc{DC}$

$\displaystyle m \overarc{AB}+ m \overarc{BC} = m \overarc{DC} + m \overarc{CB}$

By the Arc Addition Principle, 

$\displaystyle m \overarc{A C} = m \overarc{D B}$.

Both $\displaystyle \angle ADB$ and $\displaystyle \angle ACB$ are inscribed angles of the same circle which intercept the same arc; they are therefore of the same measure. The fact that $\displaystyle \overline{AD}$ is a diameter of the circle is actually irrelevant to the problem.

$\displaystyle m \overarc {NO}$ is the arithmetic mean of $\displaystyle m \overarc {NP}$ and $\displaystyle m \overarc {OP}$, so

$\displaystyle m \overarc {NO} = \frac{1}{2} \left ( m \overarc {NP} + m \overarc {OP} \right )$

By arc addition, this becomes

$\displaystyle m \overarc {NO} = \frac{1}{2} \cdot m \overarc {NPO}$

Also, $\displaystyle m \overarc {NO} + m \overarc {NPO}= 360 ^{\circ }$, or, equivalently,

$\displaystyle m \overarc {NPO} = 360 ^{\circ } - m \overarc {NO}$, so

$\displaystyle m \overarc {NO} = \frac{1}{2} \cdot \left ( 360 ^{\circ } - m \overarc {NO} \right )$

Solving for $\displaystyle m \overarc {NO}$:

$\displaystyle m \overarc {NO} = \frac{1}{2} \cdot 360 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO}$

$\displaystyle m \overarc {NO} = 180 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO}$

$\displaystyle m \overarc {NO} + \frac{1}{2} \cdot m \overarc {NO} = 180 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO} + \frac{1}{2} \cdot m \overarc {NO}$

$\displaystyle \frac{3}{2} \cdot m \overarc {NO} = 180 ^{\circ }$

$\displaystyle \frac{2} {3} \cdot \frac{3}{2} \cdot m \overarc {NO} = \frac{2} {3} \cdot 180 ^{\circ }$

$\displaystyle m \overarc {NO} = 120 ^{\circ }$

Also,

$\displaystyle m \overarc {NPO} = 360 ^{\circ } - m \overarc {NO} = 360 ^{\circ } -120 ^{\circ } = 240 ^{\circ }$

If two tangents are drawn to a circle, the measure of the angle they form is half the difference of the measures of the arcs they intercept, so

$\displaystyle m \angle NTO = \frac{1}{2}(m \overarc{NPO} - m \overarc{N O} ) = \frac{1}{2}(240^{\circ }- 120^{\circ } ) = \frac{1}{2}(120^{\circ } ) = 60^{\circ }$