Solve sin θ = -0.204 for 90º < θ < 270º. Give your answer to the nearest tenth of a degree.
Honestly have no idea how to do this.
so Ø must be in quadrant II or III , but in II the sine would be positive,
so Ø is in III
get the "angle in standard position" by using +.204
sin^-1 (.204) = 11.77°
so in III, the angle is 180 + 11.77
= 191.77°
= 191.8° correct to one decimal
check with your calculator:
sin 191.77 = -.20398.. , ok then !
To solve this equation, we need to find the value of θ that satisfies the equation sin θ = -0.204 in the given range, which is 90º < θ < 270º. Here's how you can solve it step-by-step:
1. Start by finding the reference angle: Since sin θ is negative and within the range of 90º to 270º, the reference angle will be the angle in the first quadrant that has the same absolute value as sin θ. In this case, the reference angle will be the angle whose sin is 0.204.
2. Use the inverse sine function (also known as arcsin or sin^-1) to find the reference angle: arcsin(0.204) ≈ 11.8º (rounded to one decimal place). This is the reference angle for the positive value of sin θ.
3. Since sin θ is negative, we need to find the angle in the given range that has the same reference angle as in step 2, but in the corresponding quadrant (in this case, the third quadrant).
4. Subtract the reference angle from 180º to find the angle in the third quadrant: 180º - 11.8º ≈ 168.2º (rounded to one decimal place). This is the angle in the third quadrant that has the same reference angle as in step 2.
Therefore, the solution to the equation sin θ = -0.204 in the range 90º < θ < 270º is approximately θ = 168.2º (rounded to one decimal place).