AC-DC=AD
AD = 1, this makes triangle ADC an isosceles triangle
Sum of interior opposite angles = exterior angle.
\(\angle BAD + \angle ABD = \angle BDC\)
Here these two angles are equal, lets say x.
Since AC is a straight line, \(\angle BDC = 180^ {\circ} - 120^ {\circ} = 60^ {\circ}\)
Hence \(\angle BAD + \angle ABD = 60^ {\circ}\)
x+x=60
x=30
Answer is C
AD = 1, this makes triangle ADC an isosceles triangle
Sum of interior opposite angles = exterior angle.
\(\angle BAD + \angle ABD = \angle BDC\)
Here these two angles are equal, lets say x.
Since AC is a straight line, \(\angle BDC = 180^ {\circ} - 120^ {\circ} = 60^ {\circ}\)
Hence \(\angle BAD + \angle ABD = 60^ {\circ}\)
x+x=60
x=30
Answer is C
