Question

find ( mangle a ), ( mangle b ), and ( moverarc{cd} ).( mangle a = square^circ )( mangle b = square^circ )( moverarc{cd} = square^circ )

Explanation:

Step1: Find $m\angle A$ using inscribed angle

An inscribed angle is half the measure of its intercepted arc. $\angle A$ intercepts arc $BCD$. First, calculate arc $BCD$: total circle is $360^\circ$, so $m\widehat{BCD}=360^\circ - 140^\circ=220^\circ$. Then $m\angle A=\frac{1}{2}m\widehat{BCD}$
$m\angle A=\frac{1}{2}\times220^\circ=110^\circ$

Step2: Find $m\angle B$ using inscribed angle

$\angle B$ intercepts arc $CDA$. Arc $CDA = 140^\circ + m\widehat{CD}$. First find $m\widehat{CD}$ using $\angle D$: $\angle D$ intercepts arc $ABC$, so $m\widehat{ABC}=2\times73^\circ=146^\circ$. Arc $ABC = 30^\circ + m\widehat{CD}$, so $m\widehat{CD}=146^\circ - 30^\circ=116^\circ$. Now arc $CDA=140^\circ+116^\circ=256^\circ$. Then $m\angle B=\frac{1}{2}m\widehat{CDA}$
$m\angle B=\frac{1}{2}\times256^\circ=128^\circ$

Step3: Confirm $m\widehat{CD}$

From Step2, $m\widehat{ABC}=2\times m\angle D=146^\circ$, so $m\widehat{CD}=m\widehat{ABC}-m\widehat{AB}$
$m\widehat{CD}=146^\circ - 30^\circ=116^\circ$

Answer:

$m\angle A = 110^\circ$
$m\angle B = 128^\circ$
$m\widehat{CD} = 116^\circ$