Question

please help me Q.no.19​

Answer 1

Step-by-step explanation:

$In \: \: \triangle \: ABC$

Wing angle sum property –

$∠A + ∠B + ∠C = 180°$

$40° + ∠B + ∠C = 180°$

$∠B + ∠C = 180° - 40°$

$∠B + ∠C = 1 40° - - - (i)$

$now \: In \: \: \triangle BOC$

Wing angle sum property –

$\scriptsize\angle{BOC}+ \angle{OBC} + \angle{OCB }= 180 \: (OB \: and \: OC \: bisect \angle{B} \: and \: \angle{C } \: respectively)$

$∠BOC + \frac{1}{2} ∠B + \frac{1}{2} ∠C = 180$

$\scriptsize∠BOC + \frac{1}{2} (∠B + ∠C) = 180$

$\scriptsize∠BOC = 180 - \frac{1}{2} (∠B + ∠C)$

$\scriptsize∠BOC = 180° - \frac{1}{ \cancel2}( \cancel{140°}) \: (from \: equation \: (i))$

$∠BOC = 180° - 70°$

$∠BOC = 110°$

Hence proved