Solution:
In triangle DBC
tan 600 = 10/BC
Substituting the values
√3 = 10/BC
BC = 10/√3
In triangle DBC
tan 300 = 10/ (BC + AB)
Substituting the values
1/√3 = 10/[10/√3 + AB]
By further calculation
1/√3 [10/√3 + AB] = 10
So we get
AB = 10√3 – 10/√3
Taking LCM
AB = (30 – 10)/ √3
AB = 20/√3
AB = 20√3/3
So we get
AB = (20 × 1.732)/ 3
AB = 20 × 0.577
AB = 11.540 m
Hence, the distance between A and B is 11.54 m.