The sum of the measures of the interior angles of a quadrilateral is 360 degrees. The sum of the measures of the interior angles of any polygon can be determined using the following formula:

$\displaystyle Sum \ of\ Interior \ Angles = 180^{\circ}(n-2)$, where $\displaystyle n$ is the number of sides.

For example, with a quadrilateral, which has 4 sides, you obtain the following calculation:

$\displaystyle Sum \ of\ Interior \ Angles = 180^{\circ}(4-2)=180^{\circ}\cdot2=360^{\circ}$

Solving for $\displaystyle x$ requires setting up an algebraic equation, adding all 4 angles to equal 360 degrees:

$\displaystyle 50^{\circ}+67^{\circ}+115^{\circ}+x^{\circ}=360^{\circ}$

Solving for $\displaystyle x$ is straightforward: subtract the values of the 3 known angles from both sides:

$\displaystyle x^{\circ}= 360^{\circ}-50^{\circ}-67^{\circ}-115^{\circ}$

$\displaystyle x^{\circ}=128^{\circ}$

$\displaystyle \angle AOB$ is the central angle that intercepts $\displaystyle \overarc{AB}$, so 

$\displaystyle m \angle AOB = m \overarc{AB}$.

Therefore, we need to find $\displaystyle m \overarc{AB}$ to obtain our answer.

If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore, 

$\displaystyle \frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $\displaystyle x= m \overarc{AB}$, since the total arc measure of a circle is 360 degrees, 

$\displaystyle m \overarc{ACB} = 360 - x$

We are also given that 

$\displaystyle m \angle AVB = 45$

Making substitutions, and solving for $\displaystyle x$:

$\displaystyle \frac{(360- x )- x }{2} =45$

$\displaystyle \frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\displaystyle \frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

$\displaystyle 360- 2 x = 90$

Subtract 360 from both sides:

$\displaystyle 360- 2 x - 360 = 90 - 360$

$\displaystyle -2x = -270$

Divide both sides by $\displaystyle -2$:

$\displaystyle -2x \div (-2) = -270 \div (-2)$

$\displaystyle x = 135$,

the measure of $\displaystyle \overarc{AB}$ and, consequently, that of $\displaystyle \angle AOB$. 

Consider the figure below, which adds some radii of the heptagon  (and circle):

$\displaystyle \overline{OB}$, as a radius of a regular polygon, bisects $\displaystyle \angle EBD$. The measure of this angle can be calculated using the formula

$\displaystyle m = \frac{180 (n-2)}{n}$,

where $\displaystyle n = 7$:

$\displaystyle m \angle EBD = \frac{180^{\circ } (7-2)}{7} = \frac{180^{\circ } (5)}{7} = \frac{900}{7}^{\circ } = 1 28 \frac{4}{7}^{\circ }$

Consequently,

$\displaystyle \angle DBO = \frac{1}{2} \cdot m \angle EBD = \frac{1}{2} \cdot 1 28 \frac{4}{7}^{\circ } = 64 \frac{2}{7}^{\circ }$,

the correct response.

Examine the diagram below, which divides $\displaystyle \angle AOB$ into three congruent angles, one of which is $\displaystyle \angle 1$:

The measure of a central angle of a regular $\displaystyle n$-sided polygon which intercepts one side of the polygon is $\displaystyle \frac{360 ^{\circ }}{n}$; setting $\displaystyle n = 7$, the measure of $\displaystyle \angle 1$ is 

$\displaystyle m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$. $\displaystyle \angle AOC$ has measure three times this; that is, 

$\displaystyle m \angle AOC = 3 \times51 \frac{3}{7}^{\circ } = 154 \frac{2}{7}^{\circ }$

Examine the diagram below, which divides $\displaystyle \angle AOC$ into two congruent angles, one of which is $\displaystyle \angle 1$:

The measure of a central angle of a regular $\displaystyle n$-sided polygon which intercepts one side of the polygon is $\displaystyle \frac{360 ^{\circ }}{n}$; setting $\displaystyle n = 7$, the measure of $\displaystyle \angle 1$ is 

$\displaystyle m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$. $\displaystyle \angle AOC$ has measure twice this; that is, 

$\displaystyle m \angle AOC = 2 \times51 \frac{3}{7}^{\circ } = 102 \frac{6}{7}^{\circ }$

Examine the diagram below, which divides $\displaystyle \angle AOB$ into two congruent angles, one of which is $\displaystyle \angle 1$:

The measure of a central angle of a regular $\displaystyle n$-sided polygon which intercepts one side of the polygon is $\displaystyle \frac{360 ^{\circ }}{n}$; setting $\displaystyle n = 10$, the measure of $\displaystyle \angle 1$ is 

$\displaystyle m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$. $\displaystyle \angle AOB$ has measure twice this; that is, 

$\displaystyle m \angle AOB = 2 \times 36 ^{\circ } = 72 ^{\circ }$.

Consider the triangle $\displaystyle \bigtriangleup OBC$. Since $\displaystyle \overline{OB}$ and $\displaystyle \overline{OC}$ are radii, they are congruent, and by the Isosceles Triangle Theorem, $\displaystyle m \angle OBC = m \angle OCB$. 

Now, examine the figure below, which divides $\displaystyle \angle BOC$ into three congruent angles, one of which is $\displaystyle \angle 1$:

The measure of a central angle of a regular $\displaystyle n$-sided polygon which intercepts one side of the polygon is $\displaystyle \frac{360 ^{\circ }}{n}$; setting $\displaystyle n = 10$, the measure of $\displaystyle \angle 1$ is 

$\displaystyle m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$. $\displaystyle \angle BOC$ has measure three times this; that is, 

$\displaystyle m \angle BOC = 3 \times 36 ^{\circ } = 108 ^{\circ }$.

The measures of the interior angles of a triangle total $\displaystyle 108^{\circ }$, so

$\displaystyle m \angle OCB + m \angle OBC+m \angle BOC = 180$

Substituting 108 for $\displaystyle \angle BOC$ and $\displaystyle m \angle OBC$ for $\displaystyle m \angle OCB$:

$\displaystyle m \angle OBC+ m \angle OBC+108 = 180$

$\displaystyle 2 m \angle OBC+108 = 180$

$\displaystyle 2 m \angle OBC+108- 108 = 180- 108$

$\displaystyle 2 m \angle OBC=72$

$\displaystyle 2 m \angle OBC \div 2 =72 \div 2$

$\displaystyle m \angle OBC= 36$

Through symmetry, it can be seen that Quadrilateral $\displaystyle BCDE$ is a trapezoid, such that  $\displaystyle \overline{BC} || \overline{DE }$. By the Same-Side Interior Angle Theorem, $\displaystyle \angle CBE$ and $\displaystyle \angle DEB$ are supplementary - that is, 

$\displaystyle m \angle CBE +\angle DEB = 180$.

The measure of $\displaystyle \angle DEB$ can be calculated using the formula

$\displaystyle m = \frac{180 (n-2)}{n}$,

where $\displaystyle n = 10$:

$\displaystyle m \angle DEB= \frac{180^{\circ } (10-2)}{10} = \frac{180^{\circ } (8)}{10} = \frac{1,440}{10}^{\circ } =144 ^{\circ }$

Substituting: 

$\displaystyle m \angle CBE +144= 180$

$\displaystyle m \angle CBE +144 - 144 = 180 - 144$

$\displaystyle m \angle CBE = 36$

Step 1: Define supplementary angles. Supplementary angles are two angles whose sum is $\displaystyle 180^{\circ}$.

Step 2: Find the other angle by subtracting the given angle from the maximum sum of the two angles.

So, $\displaystyle 180^{\circ}-95^{\circ}=85^\circ$

The missing angle (or second angle) is $\displaystyle 85^{\circ}$

$\displaystyle \angle 1$ and $\displaystyle \angle 3$ are supplementary angles, so, by definition,

$\displaystyle m \angle 1 + m \angle 3 = 180^{\circ }$

$\displaystyle m \angle 3= 112^{\circ }$, so substitute and solve for $\displaystyle m \angle 1$:

$\displaystyle m \angle 1 + 112^{\circ } = 180^{\circ }$

$\displaystyle m \angle 1 + 112^{\circ } -112^{\circ } = 180^{\circ }-112^{\circ }$

$\displaystyle m \angle 1 = 68 ^{\circ }$

$\displaystyle \angle 1$ and $\displaystyle \angle 2$ are complementary angles, so, by definition, 

$\displaystyle m \angle 1 + m \angle 2 = 90^{\circ }$

Substitute and solve for $\displaystyle m \angle 2$:

$\displaystyle 68 ^{\circ } + m \angle 2 = 90^{\circ }$

$\displaystyle 68 ^{\circ } + m \angle 2 - 68 ^{\circ } = 90^{\circ }-68 ^{\circ }$

$\displaystyle m \angle 2 =22^{\circ }$

$\displaystyle \angle 2 \cong \angle 4$ - that is, the angles have the same measure. Therefore,

$\displaystyle m \angle 4 =22^{\circ }$.