A stationary person observes that rain is falling vertically down at `30 km h^-1`. A cyclist is moving up on an inclined plane making an angle `30^@` with horizontal at `10 km h^-1`. In which direction should the cyclist hold his umbrella to prevent himself from the rain ?

Options

A. At an angle`tan^-1 ((3sqrt(3))/(5))` with inclined plane.
B. At an angle `tan^-1 ((3sqrt(3))/(5))` with horizontal.
C. A an angle `tan^-1 (sqrt(3)/(7))` with inclined plane.
D. At an angle `tan^-1 (sqrt(3)/(7))` with vertical.

Explanation

To solve the problem of determining the direction in which the cyclist should hold his umbrella to prevent himself from the rain, we will use the concept of relative motion. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the velocities
- The velocity of the rain is given as \( V_r = 30 \, \text{km/h} \) downward (which can be represented as \( -30 \, \hat{j} \)).
- The cyclist is moving up an inclined plane at an angle of \( 30^\circ \) with the horizontal at a speed of \( V_c = 10 \, \text{km/h} \).

### Step 2: Resolve the cyclist's velocity
- The cyclist's velocity can be resolved into two components:
- Horizontal component: \( V_{c_x} = V_c \cos(30^\circ) = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{km/h} \)
- Vertical component: \( V_{c_y} = V_c \sin(30^\circ) = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{km/h} \)

Thus, the cyclist's velocity vector is:
\[ \vec{V_c} = 5\sqrt{3} \, \hat{i} + 5 \, \hat{j} \]

### Step 3: Calculate the relative velocity of the rain with respect to the cyclist
- The relative velocity of the rain with respect to the cyclist is given by:
\[ \vec{V_{r/c}} = \vec{V_r} - \vec{V_c} \]
Substituting the values:
\[ \vec{V_{r/c}} = (-30 \, \hat{j}) - (5\sqrt{3} \, \hat{i} + 5 \, \hat{j}) = -5\sqrt{3} \, \hat{i} - 35 \, \hat{j} \]

### Step 4: Determine the direction of the relative velocity vector
- The relative velocity vector \( \vec{V_{r/c}} = -5\sqrt{3} \, \hat{i} - 35 \, \hat{j} \) indicates that the rain appears to come from a direction that is inclined.
- To find the angle \( \alpha \) that this vector makes with the vertical, we can use the tangent function:
\[ \tan(\alpha) = \frac{\text{horizontal component}}{\text{vertical component}} = \frac{5\sqrt{3}}{35} = \frac{\sqrt{3}}{7} \]

### Step 5: Find the angle with respect to the vertical
- Therefore, the angle \( \alpha \) can be calculated as:
\[ \alpha = \tan^{-1}\left(\frac{\sqrt{3}}{7}\right) \]

### Step 6: Determine the angle with respect to the incline
- Since the incline is at \( 30^\circ \), the angle \( \beta \) that the umbrella should be held with respect to the incline is:
\[ \beta = 60^\circ - \alpha \]

### Conclusion
- The cyclist should hold his umbrella at an angle \( \tan^{-1}\left(\frac{\sqrt{3}}{7}\right) \) with the vertical, and \( 60^\circ - \tan^{-1}\left(\frac{\sqrt{3}}{7}\right) \) with the inclined plane.