In figure, `angleOAB=30^(@) and angleOCB=57^(@)." Find "angleBOC and angleAOC`.

Explanation

Given,`angleOAB=30^(@) and angleOCB=57^(@)`
In `DeltaAOB, AO=OB` [both are the radius of a circle]
`rArr angleOBA=angleBAO=30^(@)`
[angles opposite to equal sides are equal]
In`DeltaAOB`,
`rArr angleAOB+angleOBA+angleBAO=180^(@)` [by angle sum property of a triangle]
`:. angle AOB+30^(@)+30^(@)=180^(@)`
`:. angleAOB=180^(@)-2(30^(@))`
`=180^(@)-60^(@)=120^(@)` ...(i)
Now, in `DeltaOCB`,
OC=OB [both are the radius of a circle]
`rArr angleOBC=angleOCB=57^(@)`
[angle opposite to equal sides are equal]
In `DeltaOCB`,
`angleCOB+angleOCB+angleCBO=180^(@)` [by angle sum property of triangle]
`:. angleCOB=180^(@)-(angleOCB+angleOBC)`
`=180^(@)-(57^(@)+57^(@))`
`=180^(@)-114^(@)=66^(@)` ...(ii)
Form Eq. (i), `angleAOB=120^(@)`
`rArr angleAOC+angleCOB=120^(@)`
`rArr angleAOC+66^(@)=120^(@)` [from Eq. (ii)]
`:. angleAOC=120^(@)-66^(@)=54^(@)`