Question

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​

Answer 1

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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