given: tanθ = √3 = √3/1 , and θ is in III
recall that tan θ = y/x
so y = -√3, and x = -1
so r^2 = x^2 + y^2 = 4
r = 2
sketch that triangle in III
then
sin θ = -√3/2
csc θ = -2/√3
Of course you might have recognized tan θ = √3 to be one of the standard
ratios from the 30-60-90° triangle with sides 1, √3, 2 respectively
and tanθ=√3 ----> θ = 60° as the relative angle
making sin θ = sin 60° = √3/2
but θ = 240° in III, so sin 240° = -√3/2
making csc θ = -2/√3
Find the exact value of csc theta is tan = sqrt 3 and the terminal side of theta is in quadrant 3.
How would I go about solving this?
4 answers
Thank you!
csc theta= -2sqrt3/3
so third option
so third option
I'm sorry, but that appears to be incorrect. Based on the given information that tan(theta) = sqrt(3) and the terminal side of theta is in quadrant 3, we can determine that:
- The opposite side (y) is negative and equal to sqrt(3)
- The adjacent side (x) is negative and equal to -1
Using the Pythagorean theorem, we can find the hypotenuse (r):
r^2 = x^2 + y^2 = (-1)^2 + (-sqrt(3))^2
r^2 = 1 + 3
r^2 = 4
r = 2
Now, we can use the definitions of trigonometric functions to find csc(theta):
csc(theta) = r / y
csc(theta) = 2 / (-sqrt(3))
csc(theta) = -2sqrt(3) / 3
Therefore, the exact value of csc(theta) is -2sqrt(3)/3. So the third option is correct.
- The opposite side (y) is negative and equal to sqrt(3)
- The adjacent side (x) is negative and equal to -1
Using the Pythagorean theorem, we can find the hypotenuse (r):
r^2 = x^2 + y^2 = (-1)^2 + (-sqrt(3))^2
r^2 = 1 + 3
r^2 = 4
r = 2
Now, we can use the definitions of trigonometric functions to find csc(theta):
csc(theta) = r / y
csc(theta) = 2 / (-sqrt(3))
csc(theta) = -2sqrt(3) / 3
Therefore, the exact value of csc(theta) is -2sqrt(3)/3. So the third option is correct.
30 answers