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?
the answer is either 0.7 or 0.07
To find the value of x, we can use the trigonometric identities and equations for a right triangle.
In a right triangle, one of the acute angles is θ and the other acute angle is Ø. Since cos Ø = x + 0.2, we can write the equation:
cos Ø = adjacent/hypotenuse = (x + 0.2)/1
Similarly, sin θ = 2x - 0.5, so we can write:
sin θ = opposite/hypotenuse = (2x - 0.5)/1
In a right triangle, the hypotenuse is always 1. Using the Pythagorean identity sin^2θ + cos^2θ = 1, we can substitute the values of sin θ and cos Ø to get:
[(2x - 0.5)/1]^2 + [(x + 0.2)/1]^2 = 1
Simplifying this equation gives:
(2x - 0.5)^2 + (x + 0.2)^2 = 1
Expanding and rearranging the terms, we get:
4x^2 - 2x + 0.25 + x^2 + 0.4x + 0.04 = 1
Combining like terms, we have:
5x^2 + 3.4x - 0.71 = 1
Rearranging the equation, we get:
5x^2 + 3.4x - 1.71 = 0
Now we can solve this quadratic equation. Factoring will not work, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c, we have:
x = (-3.4 ± √(3.4^2 - 4(5)(-1.71))) / (2(5))
Calculating the values inside the square root, we get:
x = (-3.4 ± √(11.56 + 34.2)) / 10
x = (-3.4 ± √45.76) / 10
x = (-3.4 ± 6.76) / 10
x = (-3.4 + 6.76) / 10 or x = (-3.4 - 6.76) / 10
x = 3.36/10 or x = -10.16/10
x ≈ 0.34 or x ≈ -1.02
Since x is the length of a side of the triangle, it cannot be negative. Therefore, the only valid solution is x ≈ 0.34.
Therefore, the value of x is approximately 0.34, not 0.7 or 0.07.