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:

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

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

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

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