So CD=AC=0. 1, BCD = 180-60-60 = 60,

So CB is the middle vertical line of AD at the bottom of△△ CAD, so BD=BA,

In △ABC, AB/sin∠BCA=AC/sin∠ABC.

That is ab = acsin60/sin15 = (3 √ 2+√ 6)/20.

Therefore, BD=(3√2+√6)/20≈0.33km.

Therefore, the distance between b and d is about 0.33km.