Sitting on the shore of a river, a man is in the line of a 1.0 m...

The distance of the centre of the boat from the man is $5.5 \text{ m}$ and its length is $1.0 \text{ m}$.

The nearest and farthest edges of the boat are at distances $5.5 - 0.5 = 5.0 \text{ m}$ and $5.5 + 0.5 = 6.0 \text{ m}$ respectively.

For a successful shot, the range $R$ of the projectile must satisfy $5.0 \le R \le 6.0$.

The range of a projectile is given by $R = \dfrac{v^2 \sin 2\theta}{g}$.

Substituting $v = 10 \text{ m/s}$ and $g = 10 \text{ m/s}^2$, we get:

$R = \dfrac{10^2 \sin 2\theta}{10} = 10 \sin 2\theta$

Therefore, the inequality becomes:

$5.0 \le 10 \sin 2\theta \le 6.0$

$0.5 \le \sin 2\theta \le 0.6$

The principal values for $2\theta$ are between 30^{}^{{}^{\circ}} and 37^{}^{{}^{\circ}}. Since \sin(180^{}^{{}^{\circ}} - x) = \sin x, the other possible values for $2\theta$ are between 180^{}^{{}^{\circ}} - 37^{}^{{}^{\circ}} = 143^{}^{{}^{\circ}} and 180^{}^{{}^{\circ}} - 30^{}^{{}^{\circ}} = 150^{}^{{}^{\circ}}.

Thus, the possible ranges for $2\theta$ are:

30^{}^{{}^{\circ}} \le 2\theta \le 37^{}^{{}^{\circ}} \Rightarrow 15^{}^{{}^{\circ}} \le \theta \le 18.5^{}^{{}^{\circ}}

143^{}^{{}^{\circ}} \le 2\theta \le 150^{}^{{}^{\circ}} \Rightarrow 71.5^{}^{{}^{\circ}} \le \theta \le 75^{}^{{}^{\circ}}

The minimum angle of projection is 15^{}^{{}^{\circ}} and the maximum angle is 75^{}^{{}^{\circ}}.

The forbidden interval of angles for a successful shot is between 18.5^{}^{{}^{\circ}} and 71.5^{}^{{}^{\circ}}.

The width of this forbidden interval is 71.5^{}^{{}^{\circ}} - 18.5^{}^{{}^{\circ}} = 53^{}^{{}^{\circ}}.