Question

Given: EFGH inscribed in k(O) m∠FHE = 45°, m∠EGH = 49° Find: m∠FEH

Answer 1

`m<FEH=86\°`

Explanation:

we know that

The inscribed angle measures half that of the arc comprising

step 1

Find the measure of arc EF

`m<FHE=(1)/(2)(arc\ EF)`

we have

`m<FHE=45\°`

substitute

`45\°=(1)/(2)(arc\ EF)`

`arc\ EF=90\°`

step 2

Find the measure of arc EH

`m<EGH=(1)/(2)(arc\ EH)`

we have

`m<EGH=49\°`

substitute

`49\°=(1)/(2)(arc\ EH)`

`arc\ EH=98\°`

step 3

Find the measure of arc FGH

`arc\ FGH=360\°-(arc\ EH+arc\ EF)`

substitute the values

`arc\ FGH=360\°-(98\°+90\°)`

`arc\ FGH=172\°`

step 4

Find the measure of angle FEH

`m<FEH=(1)/(2)(arc\ FGH)`

we have

`arc\ FGH=172\°`

substitute

`m<FEH=(1)/(2)(172\°)=86\°`