Question

Question no .24
Given: bo and oc bisector of angle cbe and angle bcd meet at o .angle bac=80°
To find :angle boc

Answer 1

angle CBE=angle BAC +angle ACB (By exterior angle property)   1/2angle CBE=1/2angle BAC +1/2angle ACB (dividing both sides by 1/2)   angle CBO=1/2angle BAC +1/2angle ACB (since BO is the bisector)   Similarly  angle BCO=1/2angle BAC + 1/2angle ABC   In triangle BOC  angle BCO+angleBOC+angleCBO=180 (By angle sum property)   1/2angle BAC+1/2angle ABC+1/2angle BAC+1/2angle ACB+angle BOC=180  1/2(angle BAC+angle ABC+angle BAC+angle ACB)+angle BOC=180   angle BOC=180-1/2(angle BAC+angle ABC+angle ACB+angle BAC)   angle BOC=180-1/2(180+angle BAC) (since in triangle ABC, angle BAC+angle ABC+angle ACB=180)   angle BOC=180-90-1/2angle BAC   angle BOC=90-1/2angle BAC Therefore, BOC=90-1/2 80 BOC=50

Answer 2

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