A line makes angles `alpha,beta,gamma` with co-ordinates axes. If `alpha+beta=90^(@)` then `gamma` is equal to

Options

A. `0^(@)`
B. `90^(@)`
C. `180^(@)`
D. `60^(@)`

Answer

To solve the problem step by step, we need to analyze the relationship between the angles made by a line with the coordinate axes and use trigonometric identities.

### Step-by-Step Solution:

1. **Understanding the Angles**:
The line makes angles \( \alpha \), \( \beta \), and \( \gamma \) with the x-axis, y-axis, and z-axis respectively. We are given that \( \alpha + \beta = 90^\circ \).

**Hint**: Remember that the angles with the coordinate axes are related to the direction cosines of the line.

2. **Using Direction Cosines**:
The direction cosines of the line can be represented as \( l = \cos \alpha \), \( m = \cos \beta \), and \( n = \cos \gamma \). The property of direction cosines states that:
\[ l^2 + m^2 + n^2 = 1 \]
This translates to:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

**Hint**: This equation is derived from the Pythagorean theorem applied in three dimensions.

3. **Substituting \( \beta \)**:
Since \( \alpha + \beta = 90^\circ \), we can express \( \beta \) as:
\[ \beta = 90^\circ - \alpha \]
Therefore, we can substitute this into the equation for direction cosines:
\[ \cos^2 \alpha + \cos^2(90^\circ - \alpha) + \cos^2 \gamma = 1 \]

**Hint**: Use the co-function identity: \( \cos(90^\circ - x) = \sin x \).

4. **Applying the Identity**:
Using the identity, we have:
\[ \cos^2 \alpha + \sin^2 \alpha + \cos^2 \gamma = 1 \]
According to the Pythagorean identity, \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\[ 1 + \cos^2 \gamma = 1 \]

**Hint**: Simplifying this will help isolate \( \cos^2 \gamma \).

5. **Solving for \( \cos^2 \gamma \)**:
From the equation \( 1 + \cos^2 \gamma = 1 \), we can subtract 1 from both sides:
\[ \cos^2 \gamma = 0 \]

**Hint**: Think about when the cosine of an angle equals zero.

6. **Finding \( \gamma \)**:
The equation \( \cos^2 \gamma = 0 \) implies:
\[ \cos \gamma = 0 \]
The cosine of an angle is zero at:
\[ \gamma = 90^\circ + k \cdot 180^\circ \quad (k \in \mathbb{Z}) \]
For the context of angles with respect to coordinate axes, we take:
\[ \gamma = 90^\circ \]

**Hint**: Consider the range of angles typically used in three-dimensional geometry.

### Final Answer:
Thus, the value of \( \gamma \) is \( 90^\circ \).