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

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