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

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

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

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