Let the height of the tree be h and breadth of the river be b. \( \Large In \triangle DBC, \tan 60 ^{\circ} = \frac{h}{b} \) => \( \Large h = b\sqrt{3} \) ...(i) => \( \Large In \triangle DAC, \tan 30 ^{\circ} = \frac{h}{40+b} \) => \( \Large h = \frac{40+b}{\sqrt{3}} \) ...(ii) From Eqs. (i) and (ii), we get \( \Large b\sqrt{3} = \frac{40+b}{\sqrt{3}} \) => \( \Large 2b = 40 => b = 20 m \)