Identify what angle, θ , satisfies the following conditions. sinθ=21; tanθ < 0
A
30°
B
150°
C
60°
D
300°
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1
Recognize that the equation \( \sin\theta = \frac{1}{2} \) implies that \( \theta \) could be an angle where the sine value is \( \frac{1}{2} \). Common angles with this sine value are 30° and 150°.
Recall that the sine function is positive in the first and second quadrants. Therefore, the angles 30° and 150° are potential solutions.
Consider the condition \( \tan\theta < 0 \). The tangent function is negative in the second and fourth quadrants.
Since 30° is in the first quadrant where tangent is positive, it does not satisfy \( \tan\theta < 0 \).
150° is in the second quadrant where tangent is negative, thus satisfying both conditions: \( \sin\theta = \frac{1}{2} \) and \( \tan\theta < 0 \).