Let D be a point which lies below BC angle ACD=48° angle BAC=96° SO, angle ABC=42°[AB=AC] Let angle ADC=x IN ∆ACD, by sine rule, sin 48°/AD=sin x/AC AC/ BC=sin x/ sin 48°( AD=BC) In ∆ABC, BY SINE RULE, sin 96°/BC=sin 42°/AC AC/BC=sin 42°/ sin 96° sin x/ sin 48°=sin 42°/ sin 96° So, sin x=1/2, x=30°[ 150° is not possible,then sum of angles of ∆ ADC, cannot be 180°] So, angle DAC=180°-(48°+30°)=102° #KRISHIJIVI You can choose any point D , the result will always be same