Question

HELP PLZ DUE TM!!!! 20 POINTS!!!

Answer 1

`\displaystyle\bf\\m \overset{\frown}{HE}=360-m \overset{\frown}{HL}-m \overset{\frown}{EV}-m \overset{\frown}{VL}\\m\overset{\frown}{HE}=360^o-40^o-130^o-110^o=360^o-280^o=80^o\\\\m\widehat{EYH}=m\widehat{EYV}=\frac{m \overset{\frown}{EV}-m\overset{\frown}{HE}}{2}=(130^o-80)/(2)=(50^o)/(2)=\boxed{\bf25^o}`