The first thing to do for this problem is to compute the total circumference of the circle. Notice that you were given the diameter. The proper equation is therefore:

$\displaystyle C=\pi d$

For your data, this means,

$\displaystyle C=12\pi$

Now, to compute the angle, note that you have a percentage of the total circumference, based upon your arc length:

$\displaystyle \frac{13.5}{12\pi}*360=128.9155039044336$

Rounded to the nearest hundredth, this is $\displaystyle 128.92^{\circ}$.

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, $\displaystyle \angle B$, which intercepts the semicircle $\displaystyle \overarc{ABC}$, is such an angle. Consequently, $\displaystyle m \angle B = 90 ^{\circ }$, and $\displaystyle \bigtriangleup ABC$ is a right triangle. The acute angles of $\displaystyle \bigtriangleup ABC$ are complementary, so

$\displaystyle m \angle A + m \angle C = 90 ^{\circ }$

$\displaystyle (7t+1 )+ (4t+1) = 90$

$\displaystyle 7t+4t+1+1 = 90$

$\displaystyle 11t+2 = 90$

$\displaystyle 11t+2 -2 = 90 - 2$

$\displaystyle 11t = 88$

$\displaystyle 11t\div 11 = 88 \div 11$

$\displaystyle t = 8$

The measure of inscribed $\displaystyle \angle C$ is 

$\displaystyle m \angle C = (4t+1 )^{\circ } = (4 \cdot 8+1 )^{\circ } = 33 ^{\circ }$.

An inscribed angle of a circle intercepts an arc of twice its degree measure, so

$\displaystyle m \overarc {AB} = 2 \cdot m \angle C= 2 \cdot 33^{\circ } = 66 ^{\circ }$.

$\displaystyle \overline{AB}$ is a diameter, so $\displaystyle \overarc{BC A }$ is a semicircle, and

$\displaystyle m \overarc{BC A } = 180^{\circ }$,

or, equivalently,

 $\displaystyle m \overarc{BC} + m \overarc {AC } = 180^{\circ }$

In terms of $\displaystyle t$, since $\displaystyle m \overarc {AC } = t^{\circ }$,

$\displaystyle m \overarc{BC} + t ^{\circ }= 180^{\circ }$

$\displaystyle m \overarc{BC} + t^{\circ } - t ^{\circ }= 180 ^{\circ } - t^{\circ }$

$\displaystyle m \overarc{BC} =( 180 - t)^{\circ }$

$\displaystyle \overline{NB}$ and $\displaystyle \overline{NC }$, being a secant segment and a tangent segment to a circle, respectively, intercept two arcs such that the measure of the angle that the segments form is equal to one-half the difference of the measures of the intercepted arcs - that is,

$\displaystyle \frac{1}{2} (m \overarc{BC} - m \overarc {AC }) = m \angle CNB$

Setting $\displaystyle m \overarc{BC} = \left (180 - t \right )^{\circ }$, $\displaystyle m \overarc {AC } = t^{\circ }$, and $\displaystyle m \angle CNB = 34^{\circ }$:

$\displaystyle \frac{1}{2}\left [ (180 - t) - t \right ]= 34$

$\displaystyle \frac{1}{2} (180 - 2t) = 34$

$\displaystyle \frac{1}{2} \cdot 180 -\frac{1}{2} \cdot 2t = 34$

$\displaystyle 90 - t= 34$

$\displaystyle 90-t-90 = 34 -90$

$\displaystyle -t =- 56$

$\displaystyle t = 56$

The total measure of the arcs that comprise a circle is $\displaystyle 360 ^{\circ }$, so from the above diagram,

$\displaystyle m \overarc { AC}+m \overarc { CD}+ m \overarc { DB} + m \overarc {BA } = 360 ^{\circ }$

Substituting the appropriate expression for each arc measure:

$\displaystyle t+t+3t+3t = 360$

$\displaystyle 8t = 360$

$\displaystyle 8t \div 8 = 360 \div 8$

$\displaystyle t = 45$

Therefore, 

$\displaystyle m \overarc { AC}= 45^{\circ }$ 

and 

$\displaystyle m \overarc { DB} = 3 t^{\circ } = 3 \cdot 45^{\circ } = 135^{\circ }$

The measure of the angle formed by the tangent segments $\displaystyle \overline{NB}$ and $\displaystyle \overline{ND}$, which is $\displaystyle \angle BND$, is half the difference of the measures of the arcs they intercept, so 

$\displaystyle m\angle BND = \frac{1}{2} (m \overarc {DB} - m \overarc {AC})$

Substituting:

$\displaystyle m\angle BND = \frac{1}{2} (135 ^{\circ }- 45^{\circ } )$

$\displaystyle m\angle BND = \frac{1}{2} (90^{\circ } )$

$\displaystyle m\angle BND = 45 ^{\circ }$

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $\displaystyle 180^{\circ }$.

$\displaystyle \angle B$ and $\displaystyle \angle D$ are two such angles, so 

$\displaystyle m \angle B + m \angle D = 180 ^{\circ }$

Setting $\displaystyle m \angle B = 88^{\circ }$ and $\displaystyle m \angle D =u ^{\circ }$, and solving for $\displaystyle u$:

$\displaystyle 88+ u = 180$

$\displaystyle 88+ u - 88 = 180 - 88$

$\displaystyle u= 92$,

the correct response.

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $\displaystyle 180^{\circ }$.

$\displaystyle \angle A$ and $\displaystyle \angle C$ are two such angles, so 

$\displaystyle m \angle A + m \angle C= 180 ^{\circ }$

Setting $\displaystyle m \angle A = t^{\circ }$ and $\displaystyle m \angle C = (180-t)^{\circ }$, and solving for $\displaystyle t$:

$\displaystyle t + (180-t)= 180$

$\displaystyle 180+t-t= 180$

$\displaystyle 180= 180$,

The statement turns out to be true regardless of the value of $\displaystyle t$. Therefore, without further information, the value of $\displaystyle t$ cannot be determined.

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $\displaystyle 180^{\circ }$.

$\displaystyle \angle A$ and $\displaystyle \angle C$ are two such angles, so 

$\displaystyle m \angle A + m \angle C= 180 ^{\circ }$

Setting $\displaystyle m \angle A = t^{\circ }$ and $\displaystyle m \angle C = (t+12)^{\circ }$, and solving for $\displaystyle t$:

$\displaystyle t + (t+12 )= 180$

$\displaystyle 2t +12 = 180$

$\displaystyle 2t +12 - 12 = 180 - 12$

$\displaystyle 2t = 168$

$\displaystyle 2t \div 2 = 168 \div 2$

$\displaystyle t = 84$,

the correct response.

 $\displaystyle \overline{AB}$ is a diameter, so $\displaystyle \overarc{ACB}$ is a semicircle - therefore, $\displaystyle m \overarc{ACB} = 180 ^{\circ }$. By the Arc Addition Principle,

$\displaystyle m \overarc{AC}+ m \overarc{CB} = 180 ^{\circ }$

If we let $\displaystyle t ^{\circ }= m \overarc{AC}$, then

$\displaystyle t ^{\circ }+ m \overarc{CB} = 180 ^{\circ }$,

and

$\displaystyle m \overarc{CB} = (180 -t )^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\displaystyle \overline{NC}$ and $\displaystyle \overline{NB}$ are such segments intercepting $\displaystyle \overarc{AC}$ and $\displaystyle \overarc{CB}$, it holds that

$\displaystyle \frac{1}{2}\left ( m \overarc{CB}- m \overarc{AC} \right ) = m \angle CNB$

Setting $\displaystyle m \overarc{AC} = t ^{\circ }$, $\displaystyle m \overarc{CB} = (180 -t )^{\circ }$, and $\displaystyle m \angle CNB = 28 ^{\circ }$:

$\displaystyle \frac{1}{2} [ \left ( 180-t \right ) - t ] = 28$

$\displaystyle \frac{1}{2} \left ( 180-2 t \right ) = 28$

$\displaystyle 90-t = 28$

$\displaystyle 90-t + t - 28 = 28 + t - 28$

$\displaystyle 62 = t$

$\displaystyle m \overarc{AC} =62^{\circ }$

The inscribed angle that intercepts this arc, $\displaystyle \angle CBA$, has half this measure:

$\displaystyle m \angle CBA = \frac{1}{2} \cdot m \overarc{AC} = \frac{1}{2} \cdot 62^{\circ } = 31^{\circ }$.

This is the correct response.

$\displaystyle \overline{AB}$ is a diameter, so $\displaystyle \overarc {ATB}$ is a semicircle, which has measure $\displaystyle 180 ^{\circ }$. By the Arc Addition Principle,

$\displaystyle m \overarc {AT } + m \overarc {BT} = m \overarc {ATB}$

If we let $\displaystyle t ^{\circ }= m \overarc {AT }$, then, substituting:

$\displaystyle t ^{\circ }+ m \overarc {BT} = 180 ^{\circ }$,

and

$\displaystyle m \overarc {BT} =( 180 - t ) ^{\circ }$

the ratio of the length of $\displaystyle \overarc{BT}$ to that of $\displaystyle \overarc{AT}$ is 7 to 5; this is also the ratio of their degree measures; that is,

$\displaystyle \frac{m \overarc{BT}}{m \overarc{AT}} = \frac{5}{3}$

Setting $\displaystyle m \overarc {AT } = t ^{\circ }$ and $\displaystyle m \overarc {BT} =( 180 - t ) ^{\circ }$:

$\displaystyle \frac{180-t }{t} = \frac{7}{5}$

Cross-multiply, then solve for $\displaystyle t$:

$\displaystyle 7t = 5 (180-t)$

$\displaystyle 7t = 900 - 5t$

$\displaystyle 7t + 5t = 900 - 5t + 5t$

$\displaystyle 12t = 900$

$\displaystyle \frac{12t }{12}= \frac{900}{12}$

$\displaystyle t = 75$

$\displaystyle m \overarc {AT } =75^{\circ }$, and $\displaystyle m \overarc {BT} =( 180 -75 ) ^{\circ } = 105 ^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\displaystyle \overline{NB}$ and $\displaystyle \overline{NT}$ are such segments whose angle $\displaystyle \angle TNA$ intercepts $\displaystyle \overarc{AT}$ and $\displaystyle \overarc{BT}$, it holds that:

$\displaystyle m \angle TNA = \frac{1}{2}\left ( m \overarc{BT}- m \overarc{AT} \right )$

$\displaystyle m \angle TNA = \frac{1}{2}\left ( 105 ^{\circ } - 75^{\circ } \right )$

$\displaystyle m \angle TNA = \frac{1}{2}\left ( 30 ^{\circ } \right )$

$\displaystyle m \angle TNA = 15^{\circ }$

Set the area of the circle equal to four times the circumference πr² = 4(2πr). 

Cross out both π symbols and one r on each side leaves you with r = 4(2) so r = 8 and therefore d = 16.