Question

boc = 90'
+1/2A


Prove that=boc = 90° +1/2A.

15. In A ABC, the bisectors of the interior angles ZB and ZC meet at point 0​

Answer 1

Answer:

In △BOC,

∠BOC+∠OBC+∠OCB=180 (OB and OC bisect ∠B and ∠C respectively)

∠BOC+

2

1

∠B+

2

1

∠C=180

∠BOC=180−

2

1

(∠B+∠C)

∠BOC=180−

2

1

(180−∠A)

∠BOC=180−

2

1

(180−70)

∠BOC=180−90+35

∠BOC=125

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