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)
Responses

10
10

1
1

0.07
0.07

0.7

Answer 1

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:

1. \( \cos \phi = x + 0.2 \)
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:

1. \( x = \frac{1.6 + 4.09}{10} = \frac{5.69}{10} = 0.569 \)
2. \( x = \frac{1.6 - 4.09}{10} = \frac{-2.49}{10} = -0.249 \) (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**.