I guess it is to be 300 NORTH ? (0,300)
swing 1 x = -25, y = 43.3
swing 2 x = -25+80sin35 y = 43.3+80 cos 35
so now
x = 29.9 y =109
swing 3 x =29.9-200sin10 y = 109+200cos10
so now
x = -4.83 y = 306 getting close
we must go 2.83 east and -6 south
A golf ball is to be hit 300 yards straight to make a hole in one. Having only played miniature golf, your first swing makes the ball skew 50 yards N 30 W. Trying to get back on track, the second shot goes 80 yards N 35 E. Now you are very upset with this game and you swing with anger causing the ball to go 200 yards N 10 W. At this point you realize that brut strength is not working and you decide to resort to math. What should be the components of your next swing to finally make in the hole?
2 answers
changing them to direction vectors with standard angle notation.
50 yards N30°W = 50(cos120,sin120) = (-25,43.30)
80 yards N35E = 80(cos55,sin55) = (45.89 , 65.53
200 yds N10W = 200(cos100,sin100) = (-34.73 , 196.96)
r yds at Ø = r(cosØ,sinØ) = (rcosØ,rsinØ)
adding these we should get (0,300)
rcosØ = (13.84 , -5.79)
r = √(13.84^2 + (-5.79)^2) = 15.00
tanØ = -5.79/13.84
Ø = -22.7° --> in quad IV , looking at (13.84,-5.79)
His shot should be 15 yds at E 22.7° S
50 yards N30°W = 50(cos120,sin120) = (-25,43.30)
80 yards N35E = 80(cos55,sin55) = (45.89 , 65.53
200 yds N10W = 200(cos100,sin100) = (-34.73 , 196.96)
r yds at Ø = r(cosØ,sinØ) = (rcosØ,rsinØ)
adding these we should get (0,300)
rcosØ = (13.84 , -5.79)
r = √(13.84^2 + (-5.79)^2) = 15.00
tanØ = -5.79/13.84
Ø = -22.7° --> in quad IV , looking at (13.84,-5.79)
His shot should be 15 yds at E 22.7° S