If `cos43^(@)=(x)/(sqrt(x^(2)+y^(2)))`, then the value of `tan47^(@)` is.

Options

A. `(y)/(x)`
B. `(x)/(y)`
C. `(y)/(sqrt(x^(2)+y^(2)))`
D. x

Answer

To solve the problem, we need to find the value of \( \tan 47^\circ \) given that \( \cos 43^\circ = \frac{x}{\sqrt{x^2 + y^2}} \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We know that in a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side (base) to the hypotenuse. In this case, we have:
\[ \cos 43^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{x}{\sqrt{x^2 + y^2}} \]

2. **Setting Up the Triangle**:
Let's denote:
- The angle \( \theta = 43^\circ \)
- The adjacent side (base) = \( x \)
- The hypotenuse = \( \sqrt{x^2 + y^2} \)
- The opposite side (perpendicular) = \( y \)

3. **Using the Pythagorean Theorem**:
According to the Pythagorean theorem, we have:
\[ \text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2 \]
Therefore:
\[ x^2 + y^2 = (\sqrt{x^2 + y^2})^2 \]

4. **Finding \( \tan 47^\circ \)**:
We know that:
\[ \tan 47^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{y}{x} \]

5. **Using the Complementary Angle**:
Since \( 47^\circ \) is the complement of \( 43^\circ \) (i.e., \( 90^\circ - 43^\circ = 47^\circ \)), we can use the identity:
\[ \tan(90^\circ - \theta) = \cot(\theta) \]
Hence:
\[ \tan 47^\circ = \cot 43^\circ = \frac{1}{\tan 43^\circ} \]

6. **Expressing \( \tan 43^\circ \)**:
From the triangle, we can express \( \tan 43^\circ \):
\[ \tan 43^\circ = \frac{y}{x} \]

7. **Final Calculation**:
Therefore, we have:
\[ \tan 47^\circ = \frac{x}{y} \]

### Conclusion:
Thus, the value of \( \tan 47^\circ \) is \( \frac{x}{y} \).