In AABC the external bisector of angle
B and C meet at P.If angle A = 50° and
angle C = 70° find angle BPC
А
50°
70
B
с
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Answer:
ANSWER
In △ABC,
∠ABC+∠ACB+∠BAC=180
∠ABC+70+50=180
∠ABC=60
∘
∠OCB=
2
1
(180−∠ACB)
∠OCB=
2
1
(180−50)
∠OCB=65
∘
∠OBC=
2
1
(180−∠ABC)
∠OBC=
2
1
(180−60)
∠OBC=60
∘
In △OBC,
∠OCB+∠OBC+∠BOC=180
65+60+∠BOC=180
∠BOC=180−125
∴∠BOC=55
∘
Step-by-step explanation:
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