When it is raining vertically down, to a man walking on road the velocity of rain appears to be 1.5 times his velocity. To protect himself from rain he should hold the umbrella at an angle `theta` to vertical. Then `tan theta=`

Options

A. `2/sqrt(5)`
B. `sqrt(5)/2`
C. `2/3`
D. `3/2`

Answer

To solve the problem, we need to analyze the situation involving a man walking while it is raining vertically down. The velocity of the rain appears to be 1.5 times the velocity of the man. We need to find the angle \( \theta \) at which the man should hold his umbrella and specifically calculate \( \tan \theta \).

### Step-by-Step Solution:

1. **Define Variables:**
- Let the velocity of the man with respect to the ground be \( v_m = x \) m/s.
- The velocity of the rain with respect to the ground is \( v_r = v_r \) m/s (since it is falling vertically down, we can consider it as \( v_r \)).
- The velocity of the rain with respect to the man is given as \( v_{rm} = 1.5 v_m = 1.5x \) m/s.

2. **Draw the Velocity Triangle:**
- The velocity of the rain \( v_r \) is vertical (downward).
- The velocity of the man \( v_m \) is horizontal (along the road).
- The resultant velocity of the rain with respect to the man forms a right triangle where:
- One side is \( v_m = x \) (horizontal).
- The other side is \( v_r \) (vertical).
- The hypotenuse is \( v_{rm} = 1.5x \).

3. **Use the Pythagorean Theorem:**
- According to the Pythagorean theorem, we can write:
\[ (v_{rm})^2 = (v_m)^2 + (v_r)^2 \]
- Substituting the values:
\[ (1.5x)^2 = x^2 + v_r^2 \]
- This simplifies to:
\[ 2.25x^2 = x^2 + v_r^2 \]
- Rearranging gives:
\[ v_r^2 = 2.25x^2 - x^2 = 1.25x^2 \]
- Therefore:
\[ v_r = \sqrt{1.25}x = \frac{\sqrt{5}}{2}x \]

4. **Calculate \( \sin \theta \) and \( \cos \theta \):**
- From the triangle, we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{v_m}{v_{rm}} = \frac{x}{1.5x} = \frac{1}{1.5} = \frac{2}{3} \]
- For \( \cos \theta \):
\[ \cos \theta = \frac{v_r}{v_{rm}} = \frac{\frac{\sqrt{5}}{2}x}{1.5x} = \frac{\sqrt{5}/2}{1.5} = \frac{\sqrt{5}}{3} \]

5. **Calculate \( \tan \theta \):**
- Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} \]

### Final Answer:
Thus, \( \tan \theta = \frac{2}{\sqrt{5}} \).