A vertical tree stands on one bank of a straight river. From a...

Let the width of the river be $w$ and the height of the tree be $h$.

From point $A$, the angle of elevation is $60^\circ$, so $h = w \tan 60^\circ = w\sqrt{3}$.

The person walks $40\text{ m}$ from $A$ at an angle of $30^\circ$ to the river bank, moving away from the river. The new position $B$ has a component perpendicular to the river bank equal to $40 \sin 30^\circ = 20\text{ m}$, and a component parallel to the river bank equal to $40 \cos 30^\circ = 20\sqrt{3}\text{ m}$.

The total perpendicular distance from the tree's bank to $B$ is $w + 20$. The parallel distance is $20\sqrt{3}$.

The square of the horizontal distance $OB$ from the base of the tree to $B$ is:
$OB^2 = (w + 20)^2 + (20\sqrt{3})^2 = w^2 + 40w + 400 + 1200 = w^2 + 40w + 1600$

From $B$, the angle of elevation is $45^\circ$, so $h = OB \tan 45^\circ = OB$.

Squaring both sides gives $h^2 = OB^2$. Substituting $h = w\sqrt{3}$:
$(w\sqrt{3})^2 = w^2 + 40w + 1600$
$3w^2 = w^2 + 40w + 1600$
$2w^2 - 40w - 1600 = 0$
$w^2 - 20w - 800 = 0$
$(w - 40)(w + 20) = 0$

Since the width must be positive, $w = 40\text{ m}$.

Answer: $40$