Question

what is the measure of $widehat{eab}$ in circle f?
$72^circ$
$92^circ$
$148^circ$
$200^circ$
(the circle f has inscribed polygon bcde, with $widehat{cd}=88^circ$, $angle bed=70^circ$, $angle ebc=80^circ$)

Explanation:

Step1: Recall inscribed angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc. So first find arcs intercepted by given angles.

Step2: Find arc $\widehat{BD}$

$\angle BED = 70^\circ$, so $\widehat{BD} = 2\times70^\circ = 140^\circ$

Step3: Find arc $\widehat{EC}$

$\angle EBC = 80^\circ$, so $\widehat{EC} = 2\times80^\circ = 160^\circ$

Step4: Sum known arcs

Known arcs: $\widehat{CD}=88^\circ$, $\widehat{BD}=140^\circ$, $\widehat{EC}=160^\circ$. Sum: $88^\circ + 140^\circ + 160^\circ = 388^\circ$

Step5: Calculate total excess over circle

A full circle is $360^\circ$, so the overlapping arc $\widehat{ED}$ is $388^\circ - 360^\circ = 28^\circ$

Step6: Find arc $\widehat{ED}$

$\widehat{ED} = 28^\circ$

Step7: Calculate $\widehat{EAB}$

$\widehat{EAB} = 360^\circ - \widehat{ED} - \widehat{DC} = 360^\circ - 28^\circ - 88^\circ = 244^\circ$ (correction: use $\widehat{EAB}=360^\circ - (\widehat{BD}-\widehat{BC}) - \widehat{DC}$ → alternative: $\widehat{BC}=160^\circ - \widehat{ED}-\widehat{DC}=160-28-88=44^\circ$, $\widehat{AB}=360-140-44-88=88^\circ$, $\widehat{EA}=160-88=72^\circ$, so $\widehat{EAB}=72+88=160$? No, correct method: $\widehat{EAB}=360^\circ - \widehat{EDC}$. $\widehat{EDC}=\widehat{ED}+\widehat{DC}$. $\widehat{ED}=\widehat{EC}-\widehat{DC}-\widehat{CB}$, $\widehat{CB}=2\angle CEB$ no, correct: $\widehat{EAB}=360 - (\widehat{BD} + \widehat{DC} - \widehat{BC})$. Final correct step: $\widehat{EAB}=360^\circ - (\widehat{DC} + \widehat{DE})$. $\widehat{DE}=2\angle DBE$ no, use total arcs:
$\widehat{EAB} = 360^\circ - (\widehat{BC} + \widehat{CD} + \widehat{DE})$
$\widehat{BC}=160^\circ - \widehat{BE}$, $\widehat{BE}=140^\circ - \widehat{DE}$ → simpler: $\widehat{EAB}=360 - (88 + (140-80\times2+88))$ → correct final step: $\widehat{EAB}=360^\circ - (88^\circ + (140^\circ - 160^\circ + 88^\circ))= 360 - (88+68)=204$? No, the correct approach:
$\widehat{EAB} = 360^\circ - (\widehat{ED} + \widehat{DC} + \widehat{CB})$
$\widehat{CB}=160^\circ - \widehat{ED} - \widehat{DC}$ → substitute:
$\widehat{EAB}=360 - (\widehat{ED}+88+160-\widehat{ED}-88)=360-160=200^\circ$

Answer:

200° (Option D. 200°)