only give me 12 th answer
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In quadrilateral ABCD ,
angle A + angle B + angle C + angle D = 360 ° ( sum of all inner angles of a quadrilateral is 360 ° ) angle A = x + x ( since AO bisects angle A ) angle B = y + y ( since BO bisects angle B )
Therefore , 2x + 2y + 70 ° + 50 ° = 360 ° 2 ( x + y ) + 130º = 360 ° 2 ( x + y ) = 360 ° - 130º = 230 ° x + y = 230/2 = 115 ° -------- ( i )
In triangle AOB , x + y + angle AOB = 180 ° ( Sum of all angles of a triangle is 180 ° ) 115 ° + angle AOB = 180 ° ( eq . { i } )
Therefore angle AOB = 180 ° - 115 ° = 65 °
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