Question

Type the correct answer for each.

The point (x,√7/3) in the second quadrant corresponds to angle 0 on the unit circle.

________ 0=3√7
________ 0=-√ 7/2

Answer 1

If point
`\left(x,\ (√(7))/(3)\right)` is on the unit ircle, then:


`x^2+\left( ( √(7) )/(3) \right)^2=1 \\ \\ \Rightarrow x^2+ (7)/(9) =1 \\ \\ \Rightarrow x^2=1- (7)/(9) = (2)/(9) \\ \\ \Rightarrow x= \sqrt{ (2)/(9) } = ( √(2) )/(3)`

Since, the point is in the second quadrant, x is negative.

Thus,
`(x,\ y)=\left(x,\ (√(7))/(3)\right)=\left(-( √(2) )/(3),\ (√(7))/(3)\right)`

Part A:


`3 √(7) =6\left( ( √(7) )/(2) \right) \\ \\ =6\left( ( √(7) )/(3) \cdot (3)/( √(2) ) \right) \\ \\ =6\left( ( ( √(7) )/(3) )/( ( √(2) )/(3) ) \right)=-6\left( ( ( √(7) )/(3) )/( -( √(2) )/(3) ) \right) \\ \\ =-6\left( (y)/(x) \right)=-6\tan\theta`

Therefore,
`-6\tan\theta=3√(7)`.



Part B:


`- ( √(7) )/(2) =- ( √(7) )/(3) \cdot (3)/( √(2) ) \\ \\ =- ( ( √(7) )/(3) )/( ( √(2) )/(3) ) = ( ( √(7) )/(3) )/( -( √(2) )/(3) ) = (y)/(x) \\ \\ =\tan\theta`

Therefore,
`\tan\theta=- ( √(7) )/(2)`