Alex, standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is 60°. When he retreats 30 m from the bank, he finds the angle to be 45°. The breadth of the river is

Let BC be the tree of height `h’ and A be the initial position of Alex. D is the final position of Alex.

Let the breadth of river be b m. Now Δ DBC and Δ ABC are right angled at B.

In Δ ABC, tan60° = BCAB=hb... (i)

In Δ DBC, tan45° = BCBD=h30+b... (ii)

From (i) and (ii)

3=hb and 1=h30+b ⇒3b=30+b ⇒b=303-1 ⇒b=303-1×3+13+1 ⇒b=153+1 m