Question

A right triangle has two acute angles,  θ  and  ϕ , so that  cosϕ=x+0.2  and  sinθ=2x−0.5 . What is the value of x?(1 point)Responses10100.70.7110.07

Answer 1

In a right triangle, the following relationships hold true for the acute angles \(\theta\) and \(\phi\):

1. \(\sin^2 \theta + \cos^2 \theta = 1\)
2. \(\sin^2 \phi + \cos^2 \phi = 1\)
3. 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:
1. \(x \approx \frac{4.2}{6} = 0.70\)
2. \(x \approx \frac{0.6}{6} = 0.1\)

### Conclusion
The acceptable solution that is compatible with the constraints in a triangle is:
\[
\boxed{0.7}
\]