Question

find ( mangle b ), ( mangle c ), and ( moverarc{ad} ).

Explanation:

Step1: Recall inscribed angle theorem

An inscribed angle is half its intercepted arc. For $\angle B$, it intercepts $\overset{\frown}{ADC}$. $\overset{\frown}{ADC} = \overset{\frown}{AD} + \overset{\frown}{DC} = 75^\circ + 71^\circ = 146^\circ$.
$m\angle B = \frac{1}{2} \times 146^\circ$

Step2: Calculate $m\angle B$

$m\angle B = 73^\circ$

Step3: Calculate $m\angle C$

$\angle C$ intercepts $\overset{\frown}{DAB}$. $\overset{\frown}{DAB} = \overset{\frown}{DC} + \overset{\frown}{CB} + \overset{\frown}{BA}$? No, correct: $\angle C$ intercepts $\overset{\frown}{DAB} = \overset{\frown}{AD} + \overset{\frown}{AB}$. First find $\overset{\frown}{AB}$: total circle is $360^\circ$, so $\overset{\frown}{AB} = 360^\circ - 75^\circ -71^\circ -37^\circ = 177^\circ$. Then $\overset{\frown}{DAB} = 75^\circ + 177^\circ = 252^\circ$.
$m\angle C = \frac{1}{2} \times 252^\circ = 126^\circ$

Step4: Verify with cyclic quadrilateral rule

Cyclic quadrilateral opposite angles sum to $180^\circ$. $\angle B + \angle D = 73^\circ + 107^\circ$? No, $\angle D$ is given as $75^\circ$? Wait, correction: $\angle D$ is an inscribed angle intercepting $\overset{\frown}{ABC}$. $\overset{\frown}{ABC} = 37^\circ + 177^\circ = 214^\circ$, $\frac{1}{2}\times214=107$, but given $\angle ADC=75^\circ$? No, mistake: $\angle ADC$ is part of the quadrilateral, so $\angle B + \angle D = 180^\circ$, so $\angle B = 180^\circ - 75^\circ = 105^\circ$? No, wrong rule: inscribed angle for $\angle B$ intercepts $\overset{\frown}{ADC}$, which is $\overset{\frown}{AD} + \overset{\frown}{DC}$. Wait, $\angle ADC$ intercepts $\overset{\frown}{ABC} = \overset{\frown}{AB} + \overset{\frown}{BC}$. So $\angle ADC = \frac{1}{2}(\overset{\frown}{AB} + \overset{\frown}{BC})$, so $75^\circ = \frac{1}{2}(\overset{\frown}{AB} + 37^\circ)$, so $\overset{\frown}{AB} = 2\times75 -37 = 113^\circ$. Now total arcs: $\overset{\frown}{AD} + \overset{\frown}{DC} + \overset{\frown}{CB} + \overset{\frown}{BA} = 360^\circ$. So $\overset{\frown}{AD} = 360 - 71 -37 -113 = 139^\circ$.

Step5: Correct $m\angle B$

$\angle B$ intercepts $\overset{\frown}{ADC} = \overset{\frown}{AD} + \overset{\frown}{DC} = 139 +71=210^\circ$. $m\angle B = \frac{1}{2}\times210=105^\circ$

Step6: Correct $m\angle C$

$\angle C$ intercepts $\overset{\frown}{BAD} = \overset{\frown}{BA} + \overset{\frown}{AD} =113+139=252^\circ$. $m\angle C = \frac{1}{2}\times252=126^\circ$. Now check cyclic quadrilateral: $\angle B + \angle D =105+75=180$, $\angle C + \angle A = 126 + 54=180$ (since $\angle A$ intercepts $\overset{\frown}{BCD}=71+37=108$, $\frac{1}{2}\times108=54$). Correct.

Answer:

$m\angle B = 105^\circ$
$m\angle C = 126^\circ$
$m\overset{\frown}{AD} = 139^\circ$