A man rows a boat across a river at his best speed in still water. The river flows at 4 km/hr, and the man can row at 6 km/hr in still water. If the river is 5 km wide, in what direction must he head

To land directly opposite his starting point, the man must row upstream at an angle such that the component of his velocity against the river flow cancels out the river's velocity. This forms a right-angled triangle with $v_m$ as the hypotenuse, $v_r$ as the side opposite the angle of setting off upstream, and $v_{res}$ as the side adjacent to this angle.
Let $\theta$ be the angle upstream from the direction perpendicular to the river flow.
Using trigonometry:

$$\sin \theta = \frac{v_r}{v_m}$$ $$\sin \theta = \frac{4 \text{ km/hr}}{6 \text{ km/hr}}$$ $$\sin \theta = \frac{2}{3}$$

To find the angle $\theta$:

$$\theta = \arcsin\left(\frac{2}{3}\right)$$ $$\theta \approx 41.81^\circ$$

The man must set off at an angle of $41.81^\circ$ upstream from the direction perpendicular to the river flow. If the river flows east, he must row $41.81^\circ$ west of north (assuming he wants to cross north).

The effective velocity across the river, $v_{res}$, is the component of the man's velocity perpendicular to the river flow. Using the Pythagorean theorem or the cosine of the angle:

$$v_{res} = \sqrt{v_m^2 - v_r^2}$$ $$v_{res} = \sqrt{(6 \text{ km/hr})^2 - (4 \text{ km/hr})^2}$$ $$v_{res} = \sqrt{36 \text{ km}^2/\text{hr}^2 - 16 \text{ km}^2/\text{hr}^2}$$ $$v_{res} = \sqrt{20 \text{ km}^2/\text{hr}^2}$$ $$v_{res} = \sqrt{4 \times 5} \text{ km/hr}$$ $$v_{res} = 2\sqrt{5} \text{ km/hr}$$

The time taken is the river width divided by the effective velocity across the river:

$$t = \frac{d}{v_{res}}$$ $$t = \frac{5 \text{ km}}{2\sqrt{5} \text{ km/hr}}$$

To simplify, multiply the numerator and denominator by $\sqrt{5}$:

$$t = \frac{5\sqrt{5}}{2\sqrt{5}\sqrt{5}} \text{ hr}$$ $$t = \frac{5\sqrt{5}}{2 \times 5} \text{ hr}$$ $$t = \frac{\sqrt{5}}{2} \text{ hr}$$

As a decimal:

$$t \approx \frac{2.236}{2} \text{ hr}$$ $$t \approx 1.118 \text{ hr}$$

The direction in which he must set off is $\boxed{\text{41.81}^\circ \text{ upstream from the perpendicular to the river flow}}$.
The time it will take to cross the river is $\boxed{\frac{\sqrt{5}}{2} \text{ hours}}$.