Question

in triangle abc and external bisector of angle BCA and ABC meet each other at o then find the measure of angle Boc​

Answer 1

ANSWER:

• ∠ABO=∠OBC=∠ABC2

∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2

∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,

∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BAC

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,∠ABC2+∠ACB2+∠BOC=180∘

∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,∠ABC2+∠ACB2+∠BOC=180∘⟹∠ABC+∠ACB2+∠BOC=180∘

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,∠ABC2+∠ACB2+∠BOC=180∘⟹∠ABC+∠ACB2+∠BOC=180∘⟹∠BOC=180∘−∠ABC+∠ACB2

∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,∠ABC2+∠ACB2+∠BOC=180∘⟹∠ABC+∠ACB2+∠BOC=180∘⟹∠BOC=180∘−∠ABC+∠ACB2From(1),

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,∠ABC2+∠ACB2+∠BOC=180∘⟹∠ABC+∠ACB2+∠BOC=180∘⟹∠BOC=180∘−∠ABC+∠ACB2From(1),⟹∠BOC=180∘−[90∘−∠BAC2]

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,∠ABC2+∠ACB2+∠BOC=180∘⟹∠ABC+∠ACB2+∠BOC=180∘⟹∠BOC=180∘−∠ABC+∠ACB2From(1),⟹∠BOC=180∘−[90∘−∠BAC2]⟹∠BOC=180∘−90∘+∠BAC2

• ∠ABO=∠OBC=∠ABC2∠ACO=∠OCB=∠ACB2In△ABC,∠ABC+∠ACB+∠BAC=180∘⟹∠ABC+∠ACB=180∘−∠BACDividingbothsidesby2,⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)In△BOC,∠ABC2+∠ACB2+∠BOC=180∘⟹∠ABC+∠ACB2+∠BOC=180∘⟹∠BOC=180∘−∠ABC+∠ACB2From(1),⟹∠BOC=180∘−[90∘−∠BAC2]⟹∠BOC=180∘−90∘+∠BAC2⟹∠BOC=90∘+∠BAC2