Question

A right triangle has two acute angles, θ and Ø, so that cos Ø = x + 0.2 and sin θ = 2x - 0.5. What is the value of x?

a. 0.07
b. 0.7
c. 1
d. 10

Answer 1

To find the value of \( x \), we can use the trigonometric identity in a right triangle:

\[
\sin^2(θ) + \cos^2(Ø) = 1
\]

Substituting the provided expressions:

\[
(2x - 0.5)^2 + (x + 0.2)^2 = 1
\]

Now we can expand both squares:

1. \( (2x - 0.5)^2 = 4x^2 - 2(2x)(0.5) + (0.5)^2 = 4x^2 - 2x + 0.25 \)
2. \( (x + 0.2)^2 = x^2 + 2(0.2)(x) + (0.2)^2 = x^2 + 0.4x + 0.04 \)

Now combining these:

\[
4x^2 - 2x + 0.25 + x^2 + 0.4x + 0.04 = 1
\]

Combine like terms:

\[
(4x^2 + x^2) + (-2x + 0.4x) + (0.25 + 0.04) = 1
\]

This simplifies to:

\[
5x^2 - 1.6x + 0.29 = 1
\]

Now set it to equal zero:

\[
5x^2 - 1.6x + 0.29 - 1 = 0
\]
\[
5x^2 - 1.6x - 0.71 = 0
\]

Next, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 5 \), \( b = -1.6 \), and \( c = -0.71 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-1.6)^2 - 4(5)(-0.71) = 2.56 + 14.2 = 16.76
\]

Now, applying the quadratic formula:

\[
x = \frac{-(-1.6) \pm \sqrt{16.76}}{2(5)} = \frac{1.6 \pm 4.1}{10}
\]

Calculating the two possible values:

1. \( x = \frac{1.6 + 4.1}{10} = \frac{5.7}{10} = 0.57 \)
2. \( x = \frac{1.6 - 4.1}{10} = \frac{-2.5}{10} = -0.25 \) (not a valid solution since \( x \) must be non-negative)

Since the answer options were \( 0.07, 0.7, 1, 10 \), and \( 0.57 \) is approximately \( 0.7 \) when rounded, we conclude that the correct answer is:

\[
\boxed{0.7}
\]