In a right triangle, the relationship between the angles and their sine and cosine functions can be used to establish some equations. From the given information:
- \( \cos \phi = x + 0.2 \)
- \( \sin \theta = 2x - 0.5 \)
Since we have a right triangle, the angles must satisfy the equation:
\[ \sin^2 \theta + \cos^2 \phi = 1 \]
Now we can substitute the expressions for \( \sin \theta \) and \( \cos \phi \) into the equation:
\[ (2x - 0.5)^2 + (x + 0.2)^2 = 1 \]
Expanding both squares:
\[ (2x - 0.5)^2 = 4x^2 - 2(2x)(0.5) + (0.5)^2 = 4x^2 - 2x + 0.25 \]
\[ (x + 0.2)^2 = x^2 + 2(0.2)(x) + (0.2)^2 = x^2 + 0.4x + 0.04 \]
Adding these two results together:
\[ 4x^2 - 2x + 0.25 + x^2 + 0.4x + 0.04 = 1 \]
Combining like terms yields:
\[ 5x^2 - 1.6x + 0.29 = 1 \]
Now, rearranging the equation:
\[ 5x^2 - 1.6x + 0.29 - 1 = 0 \]
\[ 5x^2 - 1.6x - 0.71 = 0 \]
Now we can use the quadratic formula to solve for \(x\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 5\), \(b = -1.6\), and \(c = -0.71\):
\[ b^2 - 4ac = (-1.6)^2 - 4 \cdot 5 \cdot (-0.71) \]
Calculating this:
\[ = 2.56 + 14.2 = 16.76 \]
Now calculate \(x\):
\[ x = \frac{1.6 \pm \sqrt{16.76}}{10} \]
Calculating \( \sqrt{16.76} \):
\[ \sqrt{16.76} \approx 4.09 \]
Now we'll find \(x\):
\[ x = \frac{1.6 \pm 4.09}{10} \]
Calculating the two possible solutions:
- \( x = \frac{1.6 + 4.09}{10} = \frac{5.69}{10} = 0.569 \)
- \( x = \frac{1.6 - 4.09}{10} = \frac{-2.49}{10} = -0.249 \) (not valid since x must be positive)
Thus, we have \(x \approx 0.569\). Among the given options (10, 1, 0.07, 0.7), the closest correct approximated value is:
0.7.