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?
To solve this problem, we can make use of the Pythagorean identity for right triangles: sin^2(θ) + cos^2(θ) = 1.
Given that sin(θ) = 2x - 0.5 and cos(Ø) = x + 0.2, we can substitute these values into the Pythagorean identity:
(2x - 0.5)^2 + (x + 0.2)^2 = 1
Expanding and simplifying:
4x^2 - 2x + 0.25 + x^2 + 0.4x + 0.04 = 1
Combining like terms:
5x^2 + 1.4x - 0.71 = 1
Moving all terms to the left side:
5x^2 + 1.4x - 1.71 = 0
At this point, we can either factor the quadratic equation or use the quadratic formula to solve for x. Since factoring is not apparent, we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Using the quadratic formula, we have:
x = (-1.4 ± √((1.4)^2 - 4(5)(-1.71))) / (2(5))
x = (-1.4 ± √(1.96 + 34.2)) / 10
x = (-1.4 ± √36.16) / 10
x = (-1.4 ± 6.01) / 10
x = -0.514 or x = 0.661
Therefore, the two possible values for x are -0.514 and 0.661.