Given that sin π¦ = β2β3
πππ tan π¦ = β3 , ππππ π¦
A. 60Β°
B. 120Β°
C. 210Β°
D. 240Β°
We know that sin π¦ = β2β3 and tan π¦ = β3
Using the fact that sin^2 π¦ + cos^2 π¦ = 1, we can find cos π¦:
sin^2 π¦ + cos^2 π¦ = 1
(-2β3)^2 + cos^2 π¦ = 1
12 + cos^2 π¦ = 1
cos^2 π¦ = -11 (this is not possible, since the square of a real number cannot be negative)
Therefore, there is no real value of π¦ that satisfies sin π¦ = β2β3 and tan π¦ = β3. This may be a mistake in the problem statement.
Yes, there is a typo.
siny = -β3/2
tany = β3
gives y in QIII, so y = 240Β°
Ah, that makes more sense! Thank you for the correction.
Given that sin π¦ = -β3/2 and tan π¦ = β3, 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 -β3/2 is y = 240Β°. Therefore, the answer is D. 240Β°.
To find the value of π¦, we can use the information given about sin π¦ and tan π¦.
First, let's find the cosine of π¦ using the Pythagorean identity:
cos^2 π¦ = 1 - sin^2 π¦
Since sin π¦ = -2β3, we have:
cos^2 π¦ = 1 - (-2β3)^2
= 1 - 4(3)
= 1 - 12
= -11
Since cosine is always between -1 and 1, this result is not possible. Therefore, there is no real value of π¦ that satisfies the given conditions.
To find the value of π¦, we can use the given information about sin π¦ and tan π¦. Let's break down the process step by step:
Step 1: Use the given value of sin π¦ = β2β3. Since sin π¦ is negative, π¦ must be in either the third or fourth quadrant of the unit circle.
Step 2: Since tan π¦ = β3, we can use the fact that tan π¦ = sin π¦ / cos π¦. Substituting the given value of sin π¦ = β2β3, we get -2β3 = (β3)/cos π¦. Simplifying further, we get -2 = 1/cos π¦.
Step 3: To solve for cos π¦, we can take the reciprocal of both sides of the equation. Doing so, we get cos π¦ = -1/2.
Step 4: Now that we have the values of sin π¦ and cos π¦, we can determine the quadrant in which π¦ lies. Since sin π¦ is negative (-2β3) and cos π¦ is negative (-1/2), π¦ must lie in the third quadrant.
Step 5: To find the angle π¦ in degrees, we can use the inverse trigonometric function. Since π¦ is in the third quadrant, we can use the inverse sine function (sinβ»ΒΉ) to find π¦. In this case, sinβ»ΒΉ (-2β3) will give us the angle in radians.
Step 6: Convert the angle from radians to degrees. To convert from radians to degrees, multiply the angle in radians by (180/Ο) or approximately 57.3.
Step 7: Calculate the value of π¦ in degrees. We have already determined in step 5 that sinβ»ΒΉ (-2β3) gives us the angle in radians. Now, multiply this angle by (180/Ο) to convert it to degrees.
By following these steps, we can find the value of π¦ in degrees and determine the correct option from the given choices.