Determine the value of $\theta$θ, an angle in standard postion, where $\sin\theta=-\frac{1}{2}$sinθ=−12 and tan $\theta=\frac{\sqrt{3}}{3}$θ=√33 , $0degrees\le\theta<360degrees$0degrees≤θ<360degrees.
6 years ago
Answered By Chris W
This question is essentially as simple as taking the inverse sine and inverse tangent in each case. Let's use the positive value first, then we have to consider in which quadrant the sine or tangent is negative or positive.
I'll write theta as x.
sin x = 1/2
x = sin-1(1/2)
x = 30 degrees
But in this case we need to find the angles where sin x = -1/2. Sine is negative in quadrants III and IV. Using our answer 30 degrees as a reference angle for these quadrants:
QIII: 180 + 30= 210 degrees
QIV: 360 - 30= 330 degrees.
tan x = sqrt(3)/3
x = tan-1(sqrt(3)/3)
x = 30 degrees
Here we're looking for quadrants where tangent is postive (quadrant I and IV). Using 30 degrees as our reference angle:
6 years ago
Answered By Chris W
This question is essentially as simple as taking the inverse sine and inverse tangent in each case. Let's use the positive value first, then we have to consider in which quadrant the sine or tangent is negative or positive.
I'll write theta as x.
sin x = 1/2
x = sin-1(1/2)
x = 30 degrees
But in this case we need to find the angles where sin x = -1/2. Sine is negative in quadrants III and IV. Using our answer 30 degrees as a reference angle for these quadrants:
QIII: 180 + 30= 210 degrees
QIV: 360 - 30= 330 degrees.
tan x = sqrt(3)/3
x = tan-1(sqrt(3)/3)
x = 30 degrees
Here we're looking for quadrants where tangent is postive (quadrant I and IV). Using 30 degrees as our reference angle:
QI: 30 degrees
Q IV: 360 - 30 = 330 degrees