If `A=130^(@)and x=sinA+cosA,` then

Options

A. `x gt0`
B. `xlt0`
C. `x=0`
D. `x ge0`

Answer

To solve the problem where \( A = 130^\circ \) and \( x = \sin A + \cos A \), we will follow these steps:

### Step 1: Substitute the value of A
We start by substituting \( A \) into the expression for \( x \):
\[ x = \sin(130^\circ) + \cos(130^\circ) \]

### Step 2: Use trigonometric identities
We can rewrite \( \sin(130^\circ) \) and \( \cos(130^\circ) \) using the sine and cosine of complementary angles:
\[ \sin(130^\circ) = \sin(90^\circ + 40^\circ) = \cos(40^\circ) \]
\[ \cos(130^\circ) = \cos(90^\circ + 40^\circ) = -\sin(40^\circ) \]

### Step 3: Substitute the identities into the equation
Now we can substitute these values back into our expression for \( x \):
\[ x = \cos(40^\circ) - \sin(40^\circ) \]

### Step 4: Analyze the expression
At this point, we can analyze the expression \( x = \cos(40^\circ) - \sin(40^\circ) \). We know that both \( \cos(40^\circ) \) and \( \sin(40^\circ) \) are positive values since \( 40^\circ \) is in the first quadrant.

### Step 5: Determine the relationship between sine and cosine
Since \( 40^\circ < 45^\circ \), we know that \( \cos(40^\circ) > \sin(40^\circ) \). Therefore:
\[ x = \cos(40^\circ) - \sin(40^\circ) > 0 \]

### Conclusion
Thus, we conclude that:
\[ x > 0 \]