If ray OC stands on line AB such that ∠AOC = ∠BOC, then find ∠BOC.

Here OP bisect ∠BOC

∴∠POC=∠BOP …………(i)

and OQ bisect ∠AOC

∴∠AOQ=∠COQ …………….(ii)

and ∠AOB=180

o

[∵AOB is a straight line]

⇒∠BOP+∠DOC+∠COQ+∠AOQ=180

o

⇒∠POC+∠POC+∠COQ+∠COQ=180

o

[using (i) & (ii)]

⇒2∠POC+2∠COQ=480

o

⇒2(∠POC+∠COQ)=180

o

⇒2∠POQ=180

o

⇒∠POQ=90

o

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step-by-step explanation:

sorry no step by step