In xy-plane, a straight line `L_1` bisects the 1st quadrant and another straight line `L_2` trisects the 2nd quadrant being closer to the axis of y . The acute angle between `L_1 and L_2` is

Options

A. `60^@`
B. `120^@`
C. `90^@`
D. `75^@`

Answer

To solve the problem of finding the acute angle between the lines \( L_1 \) and \( L_2 \) in the xy-plane, we can follow these steps:

### Step 1: Understand the lines in the coordinate system
- Line \( L_1 \) bisects the first quadrant. This means it divides the first quadrant into two equal angles. The angle between the positive x-axis and \( L_1 \) is \( 45^\circ \) because the first quadrant spans from \( 0^\circ \) to \( 90^\circ \).

### Step 2: Determine the angle for line \( L_2 \)
- Line \( L_2 \) trisects the second quadrant and is closer to the y-axis. The second quadrant spans from \( 90^\circ \) to \( 180^\circ \).
- To trisect this quadrant, we divide the \( 90^\circ \) angle (from \( 90^\circ \) to \( 180^\circ \)) into three equal parts. Each part will be \( 30^\circ \).
- Therefore, the angles for \( L_2 \) will be \( 90^\circ + 30^\circ = 120^\circ \) (the first trisection) and \( 90^\circ + 60^\circ = 150^\circ \) (the second trisection). Since \( L_2 \) is closer to the y-axis, we take \( L_2 \) as \( 120^\circ \).

### Step 3: Calculate the acute angle between \( L_1 \) and \( L_2 \)
- The angle between \( L_1 \) (which is at \( 45^\circ \)) and \( L_2 \) (which is at \( 120^\circ \)) can be calculated as follows:
\[ \text{Angle between } L_1 \text{ and } L_2 = |120^\circ - 45^\circ| = 75^\circ \]

### Conclusion
- Since \( 75^\circ \) is less than \( 90^\circ \), it is the acute angle between the two lines.

Thus, the acute angle between \( L_1 \) and \( L_2 \) is \( 75^\circ \).

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