We know that sin 𝑦 = −2√3 and tan 𝑦 = √3
Using the fact that sin^2 𝑦 + cos^2 𝑦 = 1, we can find cos 𝑦:
sin^2 𝑦 + cos^2 𝑦 = 1
(-2√3)^2 + cos^2 𝑦 = 1
12 + cos^2 𝑦 = 1
cos^2 𝑦 = -11 (this is not possible, since the square of a real number cannot be negative)
Therefore, there is no real value of 𝑦 that satisfies sin 𝑦 = −2√3 and tan 𝑦 = √3. This may be a mistake in the problem statement.
Given that sin 𝑦 = −2√3
𝑎𝑛𝑑 tan 𝑦 = √3 , 𝑓𝑖𝑛𝑑 𝑦
A. 60°
B. 120°
C. 210°
D. 240°
3 answers
Yes, there is a typo.
siny = -√3/2
tany = √3
gives y in QIII, so y = 240°
siny = -√3/2
tany = √3
gives y in QIII, so y = 240°
Ah, that makes more sense! Thank you for the correction.
Given that sin 𝑦 = -√3/2 and tan 𝑦 = √3, we know that y is in the third quadrant, where sine is negative and tangent is positive.
Using the unit circle, we can see that the only angle in the third quadrant with sine equal to -√3/2 is y = 240°. Therefore, the answer is D. 240°.
Given that sin 𝑦 = -√3/2 and tan 𝑦 = √3, we know that y is in the third quadrant, where sine is negative and tangent is positive.
Using the unit circle, we can see that the only angle in the third quadrant with sine equal to -√3/2 is y = 240°. Therefore, the answer is D. 240°.