Question

fig BO and CO are the bisector of angle ABC and angle ACO then find the angle BOC if angle A = 60 and angle ACB = 50°​

Answer 1

Hope this answer helps you

Answer 2

Answer:

In △ABC,

$\sf \: ∠ABC+∠ACB+∠BAC=180$

$\sf∠ABC+70+50=180$

$\sf \: ∠ABC=60°$

$\sf \: ∠OCB= \frac{1}{2} (180−∠ACB)$

$\sf \: ∠OCB= \frac{1}{2} (180−50)$

$∠OCB=65 {}^{0}$

$\sf \: ∠OBC= \frac{1}{2} (180−∠ABC)$

$\sf \: ∠OBC= \frac{1}{2} (180−60)$

$\sf \: ∠OBC=60°$

$\sf \: In △OBC,$

$\sf \: ∠OCB+∠OBC+∠BOC=180$

$\sf \: 65+60+∠BOC=180$

$\sf \: ∠BOC=180−125$

$∴\sf ∠BOC=55°$

Hence ∠BOC=55°