At what point does the terminal side of the angle 5π6 in standard position intersect the unit circle? (-3√2, 12) (-12, 3√2) (12, −3√2) (3√2, −12) which point on the unit circle corresponds to −π6 ? (-12, 3√2) (3√2, −12) (12, −3√2) (-3√2, 12) what is the exact value of sin(5π4) ? −12 2√2 12 −2√2 what is the exact value of tan(−3π4) ? enter your answer in the box. in which quadrant does θ lie given that sinθ< 0 and cosθ> 0 ? quadrant i quadrant ii quadrant iii quadrant iv

1.) At what point does the terminal side of the angle 5π/6 in standard position intersect the unit circle?

First option: ( -√3/2, 1/2)

2. Which point on the unit circle corresponds to −π/6 ?

Second option: ( √3/2, -1/2)

3. What is the exact value of sin(5π/4) ?

Fourth option: −√2/2

4. What is the exact value of tan(−3π/4) ?

tan(−3π/4) = 1

5. In which quadrant does θ lie given that sinθ<0 and cosθ>0 ?

Fourth option: Quadrant IV

Step-by-step explanation:

1.) At what point does the terminal side of the angle 5π6 in standard position intersect the unit circle?

Point: P=(x,y)=?

x=cos 5π/6 → x = -√3/2

y=sin 5π/6 → y = 1/2

P=( -√3/2, 1/2)

2. Which point on the unit circle corresponds to −π/6 ?

Point: P=(x,y)=?

x=cos (-π/6) → x = √3/2

y=sin (-π/6) → y = -1/2

P=( √3/2, -1/2)

3. What is the exact value of sin(5π/4) ?

sin(5π/4) = −√2/2

4. What is the exact value of tan(−3π/4) ?

tan(−3π/4) = 1

5. In which quadrant does θ lie given that sinθ<0 and cosθ>0 ?

sinθ<0 (-) in Quadrants III and IV

cosθ>0 (+) in Quadrants I and IV

Then, sinθ<0 and cosθ>0 in Quadrant IV

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

true? is that what you're asking or do you mean something else?