In ∆ ABC bisectors of exterior angles B and C meet at O. Given ∠BAC= 60° and ∠ABC = 70°, find ∠BOC.
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Step-by-step explanation:
Answer
In In △ABC,
In △ABC,∠ABC+∠ACB+∠BAC=180
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB=
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 2
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB=
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 2
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC=
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 2
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC=
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 2
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘ In △OBC,
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘ In △OBC,∠OCB+∠OBC+∠BOC=180
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘ In △OBC,∠OCB+∠OBC+∠BOC=18065+60+∠BOC=180
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘ In △OBC,∠OCB+∠OBC+∠BOC=18065+60+∠BOC=180∠BOC=180−125
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘ In △OBC,∠OCB+∠OBC+∠BOC=18065+60+∠BOC=180∠BOC=180−125∴∠BOC=55
In △ABC,∠ABC+∠ACB+∠BAC=180∠ABC+70+50=180∠ABC=60 ∘ ∠OCB= 21 (180−∠ACB)∠OCB= 21 (180−50)∠OCB=65 ∘ ∠OBC= 21 (180−∠ABC)∠OBC= 21 (180−60)∠OBC=60 ∘ In △OBC,∠OCB+∠OBC+∠BOC=18065+60+∠BOC=180∠BOC=180−125∴∠BOC=55 ∘