Question
a right triangle has two angles,0 with a horizontal line in it, and o with a / going through it, so that cos o with a / going through it = x + 0.2 and sin 0 with a horizontal line going through it = 2x - 0.5. What is the value of x?
0.7
1
0.07
10
0.7
1
0.07
10
Answers
GPT-4o mini
To solve for \( x \) given the equations \( \cos \overline{o} = x + 0.2 \) and \( \sin \overline{0} = 2x - 0.5 \), we will use the Pythagorean identity for sine and cosine:
\[
\cos^2 \overline{o} + \sin^2 \overline{o} = 1
\]
We have:
\[
\cos \overline{o} = x + 0.2
\]
\[
\sin \overline{o} = 2x - 0.5
\]
Substituting these into the Pythagorean identity gives us:
\[
(x + 0.2)^2 + (2x - 0.5)^2 = 1
\]
Now we can expand both squares:
\[
(x + 0.2)^2 = x^2 + 2(0.2)x + (0.2)^2 = x^2 + 0.4x + 0.04
\]
\[
(2x - 0.5)^2 = (2x)^2 - 2(2x)(0.5) + (0.5)^2 = 4x^2 - 2x + 0.25
\]
Now substitute these into the equation:
\[
x^2 + 0.4x + 0.04 + 4x^2 - 2x + 0.25 = 1
\]
Combine the terms:
\[
(1 + 4)x^2 + (0.4 - 2)x + (0.04 + 0.25) = 1
\]
This simplifies to:
\[
5x^2 - 1.6x + 0.29 = 1
\]
Now, move 1 to the left side of the equation:
\[
5x^2 - 1.6x + 0.29 - 1 = 0
\]
This simplifies to:
\[
5x^2 - 1.6x - 0.71 = 0
\]
Now we can apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this case, \( a = 5 \), \( b = -1.6 \), and \( c = -0.71 \):
\[
x = \frac{-(-1.6) \pm \sqrt{(-1.6)^2 - 4 \cdot 5 \cdot (-0.71)}}{2 \cdot 5}
\]
Calculate the discriminant:
\[
(-1.6)^2 = 2.56
\]
\[
4 \cdot 5 \cdot (-0.71) = -14.2
\]
Thus,
\[
b^2 - 4ac = 2.56 + 14.2 = 16.76
\]
Take the square root:
\[
\sqrt{16.76} \approx 4.1
\]
Now substitute back into the quadratic formula:
\[
x = \frac{1.6 \pm 4.1}{10}
\]
This gives two solutions:
1. \( x \approx \frac{5.7}{10} = 0.57 \)
2. \( x \approx \frac{-2.5}{10} = -0.25 \)
Given the options provided \( (0.7, 1, 0.07, 10) \):
The solutions don't exactly match the options, but the closest approximate value would be \( 0.7 \).
Therefore, one of the potential values for \( x \) rounded and matching the implications of sine and cosine would be:
**0.7**
\[
\cos^2 \overline{o} + \sin^2 \overline{o} = 1
\]
We have:
\[
\cos \overline{o} = x + 0.2
\]
\[
\sin \overline{o} = 2x - 0.5
\]
Substituting these into the Pythagorean identity gives us:
\[
(x + 0.2)^2 + (2x - 0.5)^2 = 1
\]
Now we can expand both squares:
\[
(x + 0.2)^2 = x^2 + 2(0.2)x + (0.2)^2 = x^2 + 0.4x + 0.04
\]
\[
(2x - 0.5)^2 = (2x)^2 - 2(2x)(0.5) + (0.5)^2 = 4x^2 - 2x + 0.25
\]
Now substitute these into the equation:
\[
x^2 + 0.4x + 0.04 + 4x^2 - 2x + 0.25 = 1
\]
Combine the terms:
\[
(1 + 4)x^2 + (0.4 - 2)x + (0.04 + 0.25) = 1
\]
This simplifies to:
\[
5x^2 - 1.6x + 0.29 = 1
\]
Now, move 1 to the left side of the equation:
\[
5x^2 - 1.6x + 0.29 - 1 = 0
\]
This simplifies to:
\[
5x^2 - 1.6x - 0.71 = 0
\]
Now we can apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In this case, \( a = 5 \), \( b = -1.6 \), and \( c = -0.71 \):
\[
x = \frac{-(-1.6) \pm \sqrt{(-1.6)^2 - 4 \cdot 5 \cdot (-0.71)}}{2 \cdot 5}
\]
Calculate the discriminant:
\[
(-1.6)^2 = 2.56
\]
\[
4 \cdot 5 \cdot (-0.71) = -14.2
\]
Thus,
\[
b^2 - 4ac = 2.56 + 14.2 = 16.76
\]
Take the square root:
\[
\sqrt{16.76} \approx 4.1
\]
Now substitute back into the quadratic formula:
\[
x = \frac{1.6 \pm 4.1}{10}
\]
This gives two solutions:
1. \( x \approx \frac{5.7}{10} = 0.57 \)
2. \( x \approx \frac{-2.5}{10} = -0.25 \)
Given the options provided \( (0.7, 1, 0.07, 10) \):
The solutions don't exactly match the options, but the closest approximate value would be \( 0.7 \).
Therefore, one of the potential values for \( x \) rounded and matching the implications of sine and cosine would be:
**0.7**