From the given data, the fig. is made
Let width of river = PQ = (x + y) m
Height of tree (AB) = 20 m
So, in ΔABP
tan 60o = AB/ BP
√3 = 20/ x
x = 20/ √3 m
In ΔABQ,
tan 30o = AB/ BQ
1/ √3 = 20/ y
y = 20√3
So, (x + y) = 20/ √3 + 20√3
= [20 + 20(3)]/ √3
= 80/√3
Therefore, the width of the river is 80/√3 m.