A child in danger of drowning in a river is being carried downstream by a current that flows uniformly at a speed of `2.5 km//h.` The child is 0.6 km from shore and 0.8 km upstream of a boat landing when a rescue boat sets out. If the boat proceeds at its maximum speed of `20 km//h` with respect to the water, what angle does the boat velocity v make with the shore? How long will it take boat to reach the child?

Explanation

To solve the problem, we need to analyze the motion of the boat and the child in the river. Here's the step-by-step solution:

### Step 1: Define the Problem
The child is being carried downstream by a river current at a speed of 2.5 km/h. The child is 0.6 km from the shore and 0.8 km upstream from the boat landing. The boat can travel at a speed of 20 km/h relative to the water. We need to find the angle θ that the boat's velocity makes with the shore and the time it takes for the boat to reach the child.

### Step 2: Set Up the Coordinate System
- Let the x-axis be along the direction of the river current (downstream).
- Let the y-axis be perpendicular to the river current (across the river).

### Step 3: Break Down the Boat's Velocity
The boat's velocity can be broken down into two components:
- The x-component (downstream): \( v_x = 20 \cos \theta \)
- The y-component (across the river): \( v_y = 20 \sin \theta \)

### Step 4: Set Up the Equations
The boat must cover:
- A horizontal distance of 0.8 km upstream (against the current).
- A vertical distance of 0.6 km across the river.

The equations for the distances covered by the boat can be written as:
1. \( 20 \cos \theta \cdot t = 0.8 + 2.5t \) (since the current carries the child downstream)
2. \( 20 \sin \theta \cdot t = 0.6 \)

### Step 5: Solve for Time
From the second equation, we can express time \( t \):
\[ t = \frac{0.6}{20 \sin \theta} \]

### Step 6: Substitute Time into the First Equation
Substituting \( t \) into the first equation:
\[ 20 \cos \theta \cdot \left(\frac{0.6}{20 \sin \theta}\right) = 0.8 + 2.5 \left(\frac{0.6}{20 \sin \theta}\right) \]

### Step 7: Simplify the Equation
This simplifies to:
\[ \frac{0.6 \cos \theta}{\sin \theta} = 0.8 + \frac{1.5}{\sin \theta} \]
\[ 0.6 \cot \theta = 0.8 + 1.5 \]

### Step 8: Solve for \( \tan \theta \)
Rearranging gives:
\[ 0.6 \cot \theta = 2.3 \]
\[ \cot \theta = \frac{2.3}{0.6} \]
\[ \tan \theta = \frac{0.6}{2.3} = \frac{3}{4} \]

### Step 9: Find \( \sin \theta \) and \( \cos \theta \)
Using the Pythagorean identity:
- \( \sin^2 \theta + \cos^2 \theta = 1 \)
- From \( \tan \theta = \frac{3}{4} \), we can find:
- \( \sin \theta = \frac{3}{5} \)
- \( \cos \theta = \frac{4}{5} \)

### Step 10: Calculate Time Taken
Now substituting \( \sin \theta \) back into the time equation:
\[ t = \frac{0.6}{20 \cdot \frac{3}{5}} = \frac{0.6}{12} = 0.05 \text{ hours} \]
Converting hours to minutes:
\[ 0.05 \times 60 = 3 \text{ minutes} \]

### Final Answers
- The angle \( \theta \) that the boat's velocity makes with the shore is such that \( \tan \theta = \frac{3}{4} \).
- The time taken for the boat to reach the child is 3 minutes.