Answer:
Part A: α = arc tan (y/x) = tan⁻¹ (y/x)
Part B: quadrant II α = arc tan (- y/x) = arc (- tan (y/x)) = - tan⁻¹ (y/x) 180°>α>90°
quadrant III α = arc tan (-y)/(-x) = arc tan (y/x) = tan⁻¹ (y/x) 270°>α>180°
quadrant IV α = arc tan (- y/x) = arc (- tan (y/x)) = - tan⁻¹ (y/x) 360°>α>270°
Part C: quadrant II 180°>α>90° α = tan⁻¹ (-6/1) = - tan⁻¹ 6 = 180° - 80.53° = 99.47°
Step-by-step explanation:
PART A:
In this case we will use trigonometric function tanα to calculate angle α:
tanα = y/x => α = arc tan (y/x) = tan⁻¹ (y/x)
This formula is use in general way and in first quadrant 90°>α>0°
Part B:
But in the other quadrants you must know to use unit circle to reduce angle from II, III and IV quadrant to the first quadrant.
If angle is in the quadrant II 180°>α>90° then
tanα = - tan (180°-α)
If angle is in the quadrant III 270°>α>180° then
tanα = tan (α-180°)
If angle is in the quadrant IV 360°>α>270° then
tanα =- tan (360°-α)
Part C:
vector w (x,y) = (-1,6) this vector lies in quadrant II 180°>α>90°
180°- α = tan⁻¹ (-6/1) = - tan⁻¹ 6 = 80.53° => 180° - α = 80.53°
α = 180° - 80.53 = 99.47°
It's not easy to understand this, but it's not easy for me to explain.
God with you!!!
