Question

Please help! I've been working on this for a few days and I just don't understand, it's due in a few hours. Thank you.

Answer 1

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!!!