In a right triangle, the following relationships hold true for the acute angles \(\theta\) and \(\phi\):
- \(\sin^2 \theta + \cos^2 \theta = 1\)
- \(\sin^2 \phi + \cos^2 \phi = 1\)
- The sum of angles in a triangle is \(180^\circ\).
Given:
- \(\cos \phi = x + 0.2\)
- \(\sin \theta = 2x - 0.5\)
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we also have \(\sin^2 \phi + \cos^2 \phi = 1\).
Step 1: Express \(\cos \theta\) in terms of \(x\)
We have: \[ \sin^2 \theta + \cos^2 \theta = 1 \] So, \[ \cos^2 \theta = 1 - \sin^2 \theta \] Substituting \(\sin \theta = 2x - 0.5\): \[ \cos^2 \theta = 1 - (2x - 0.5)^2 \]
Step 2: Express \(\sin \phi\)
Using \(\sin^2 \phi + \cos^2 \phi = 1\): \[ \sin^2 \phi = 1 - \cos^2 \phi \] Substituting \(\cos \phi = x + 0.2\): \[ \sin^2 \phi = 1 - (x + 0.2)^2 \]
Step 3: We can use the property of triangles
From the property of right triangles, \[ \sin^2 \theta + \sin^2 \phi = 1 \]
Setting the equation: \[ (2x - 0.5)^2 + (1 - (x + 0.2)^2) = 1 \]
Step 4: Substitute and simplify
We can replace \(1 - (x + 0.2)^2\): \[ (2x - 0.5)^2 + 1 - (x + 0.2)^2 = 1 \] This simplifies to: \[ (2x - 0.5)^2 = (x + 0.2)^2 \]
Step 5: Expand both sides
Expanding the left side: \[ (2x - 0.5)(2x - 0.5) = 4x^2 - 2x + 0.25 \]
And the right side: \[ (x + 0.2)(x + 0.2) = x^2 + 0.4x + 0.04 \]
Setting the expanded equations equal: \[ 4x^2 - 2x + 0.25 = x^2 + 0.4x + 0.04 \]
Step 6: Rearrange to solve for \(x\)
Moving all terms to one side: \[ 4x^2 - x^2 - 2x - 0.4x + 0.25 - 0.04 = 0 \] This simplifies to: \[ 3x^2 - 2.4x + 0.21 = 0 \]
Step 7: Use the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2.4\), and \(c = 0.21\): \[ b^2 - 4ac = (-2.4)^2 - 4 \times 3 \times 0.21 \] Calculating: \[ 5.76 - 2.52 = 3.24 \] Now: \[ x = \frac{2.4 \pm \sqrt{3.24}}{6} \]
Calculating \(\sqrt{3.24} \approx 1.8\): \[ x = \frac{2.4 \pm 1.8}{6} \] This gives two solutions:
- \(x \approx \frac{4.2}{6} = 0.70\)
- \(x \approx \frac{0.6}{6} = 0.1\)
Conclusion
The acceptable solution that is compatible with the constraints in a triangle is: \[ \boxed{0.7} \]