Question

6. In Fig. 10.170, <OAB = 30° and <OCB = 57°. Find <BOC and <AOC.

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Answer 1

☃︎$\rm\bf\underline{Solution:}$☃︎

in figure,

OA=OB=OC

so,∠OCB=∠OBC

also,∠OBA is 30°

so, ∠ABC=27°

similarly, ∠BAC=27°

⇒∠BAC=∠ABC

⇒AC=BC

⇒∠AOC=∠BOC                       ....(1)

now by ASP in ΔBOA,

∠BOA=120°

∴∠BOC=∠AOC=60°                 (using (1) ).

Answer 2

Answer:

☃︎Solution:‾\

☃︎ figure,OA=OB=OCso,∠OCB=∠OBCalso,∠OBA is 30°so, ∠ABC=27°similarly, ∠BAC=27°⇒∠BAC=∠ABC⇒AC=BC⇒∠AOC=∠BOC ....(1)now by ASP in ΔBOA,∠BOA=120°∴∠BOC=∠AOC=60° (using (1) ).