Question

solve the following question

Answer 1

in ΔABC

∠BAC+∠ABC+∠ACB= 180°···············(eqn 1)

AND IN ΔOBC

∠OBC+∠OCB+∠BOC= 180°·················(eqn 2)

from eqn 1

∠BAC+∠ABC+∠ACB= 180°

⇒∠A+2∠OBC+2∠OCB=180°

⇒∠A+2{∠OBC+∠OCB}=180°

⇒∠A+2{180°-∠BOC}=180°      [FROM EQN 2]

⇒1/2∠A+180°-∠BOC=90°      [dividing both sides by 2]

⇒1/2∠A+90°-∠BOC=0

⇒∠BOC=1/2∠A+90°