Question
a car leaves P and moves 20km north to Q from Q it moves 15km on abearing N45 E TO R
Answers
oobleck
use the law of cosines to find the distance PR
If P is at (0,0) the R is at (x,y) = (15/√2,20+15/√2)
So the bearing of R from P is 90-θ, where tanθ = y/x
If P is at (0,0) the R is at (x,y) = (15/√2,20+15/√2)
So the bearing of R from P is 90-θ, where tanθ = y/x
henry2,
All angles are measured CW from +y-axis.
PR = PQ+QR = 20km[0o]+15km[45o]
D = (20*sin0+15*sin45)+(20*cos0+15*cos45)i
D = 10.61+30.61i = 32.4km[19.1o].
PR = PQ+QR = 20km[0o]+15km[45o]
D = (20*sin0+15*sin45)+(20*cos0+15*cos45)i
D = 10.61+30.61i = 32.4km[19.1o].