see the attached figure to better understand the problem
Let
A(-3,1) B(0,0) C(-3,0)
we know that
the point A belong to the II quadrant
so
sin θ is positive
cos θ is negative
tan θ is negative
θ + α= 180 --------> by supplementary angles
in the right triangle ABC
sin α=opposite side angle α/hypotenuse
opposite side angle α=AC------> 1 units
hypotenuse----> AB------> applying the Pythagorean Theorem
AB²=AC²+BC²------> AB²=1²+3²-----> AB=√10 units
sin α=1/√10
cos α=adjacent side angle α/hypotenuse
adjacent side angle α=BC------> 3 units
hypotenuse----> AB------> √10 units
cos α=3/√10
tan α=opposite side angle α/adjacent side angle α
opposite side angle α=AC------> 1 units
adjacent side angle α=BC------> 3 units
tan α=1/3
Find angle α
α=arc tan (1/3)------>α=18.43°
Find angle θ
θ=180°-α------> θ=180-18.43-------> θ=161.57°
Hence
the answer is
θ=161.57°
α=18.43°
sin θ=+1/√10
cos θ=-(3/√10)
tan θ=-1/3
