A car `A` going north-east at `80km//h` and another car `B` is going south-east at `60km//h.` The direction of the velocity of `A` relative to `B` makes an angle with the north equal to:

Options

A. `tan^(-1)(2//7)`
B. `tan^(-1)(7//2)`
C. `tan^(-1)(7)`
D. `tan^(-1)(1//7)`

Explanation

To solve the problem, we need to find the angle that the velocity of car A relative to car B makes with the north direction. Here’s how we can approach this step-by-step:

### Step 1: Define the velocities of both cars
- Car A is moving north-east at a speed of 80 km/h.
- Car B is moving south-east at a speed of 60 km/h.

### Step 2: Break down the velocities into components
1. **Velocity of Car A (V_A)**:
- The north-east direction means both the x (east) and y (north) components are equal.
- \( V_{A_x} = 80 \cos(45^\circ) \)
- \( V_{A_y} = 80 \sin(45^\circ) \)

Since \( \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
- \( V_{A_x} = 80 \cdot \frac{1}{\sqrt{2}} = 40\sqrt{2} \, \text{km/h} \)
- \( V_{A_y} = 80 \cdot \frac{1}{\sqrt{2}} = 40\sqrt{2} \, \text{km/h} \)

Thus, the velocity vector of A is:
\[ \vec{V_A} = 40\sqrt{2} \hat{i} + 40\sqrt{2} \hat{j} \]

2. **Velocity of Car B (V_B)**:
- The south-east direction means the x (east) component is positive and the y (south) component is negative.
- \( V_{B_x} = 60 \cos(45^\circ) \)
- \( V_{B_y} = -60 \sin(45^\circ) \)

Similarly:
- \( V_{B_x} = 60 \cdot \frac{1}{\sqrt{2}} = 30\sqrt{2} \, \text{km/h} \)
- \( V_{B_y} = -60 \cdot \frac{1}{\sqrt{2}} = -30\sqrt{2} \, \text{km/h} \)

Thus, the velocity vector of B is:
\[ \vec{V_B} = 30\sqrt{2} \hat{i} - 30\sqrt{2} \hat{j} \]

### Step 3: Calculate the relative velocity of A with respect to B
The relative velocity \( \vec{V_{AB}} \) is given by:
\[ \vec{V_{AB}} = \vec{V_A} - \vec{V_B} \]
Substituting the components:
\[ \vec{V_{AB}} = (40\sqrt{2} \hat{i} + 40\sqrt{2} \hat{j}) - (30\sqrt{2} \hat{i} - 30\sqrt{2} \hat{j}) \]
\[ \vec{V_{AB}} = (40\sqrt{2} - 30\sqrt{2}) \hat{i} + (40\sqrt{2} + 30\sqrt{2}) \hat{j} \]
\[ \vec{V_{AB}} = 10\sqrt{2} \hat{i} + 70\sqrt{2} \hat{j} \]

### Step 4: Find the angle with respect to the north direction
To find the angle \( \alpha \) that \( \vec{V_{AB}} \) makes with the north direction (y-axis), we use the tangent function:
\[ \tan(\alpha) = \frac{V_{AB_x}}{V_{AB_y}} = \frac{10\sqrt{2}}{70\sqrt{2}} = \frac{10}{70} = \frac{1}{7} \]

### Step 5: Calculate the angle
Now, we can find \( \alpha \):
\[ \alpha = \tan^{-1}\left(\frac{1}{7}\right) \]

### Conclusion
The angle that the velocity of car A relative to car B makes with the north is \( \tan^{-1}\left(\frac{1}{7}\right) \).

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