Calculate `angleAOC, angleBOD` and `angleAOE` in the adjoining figure, it is being given that `angleCOD=90^(@), angleBOE=72^(@)` and `AOB` is a straight line.

Explanation

Since AOB is a straight line, the sum of all the angles on the lower side of AOB at a point O on it, is `180^(@)`.
`:. angleAOE+angleBOE=180^(@)`
`implies 3x+72=180`
`implies 3x=(180-72)=108 implies x=108/3=36`.
Again, AOB is a straight line and O is a point on it. So, the sum of all angles on the upper side of AOB at a point O on it, is `180^(@)`.
`:. angleAOC+angleCOD+angleDOB=180^(@)`
`implies x+90+y=180`
`[ :' angleAOC=x^(@), angleCOD=90^(@) and angleDOB=y^(@)]`
`implies 36+90+y=180" "[ :' x=36]`
`implies 126+y=180 implies y=(180-126) =54`
`:. angleAOC =x^(@)=36^(@), angleBOD=y^(@)=54^(@)`,
`angleAOE=(3x)^(@)=(3xx36)^(@)=108^(@)`.

`/_AOC =36^@, /_BOD = 54^@ and /_AOE = 108^@`