X = 8 Mi
Y = -10 Mi
a. Tan A = Y/X = -10/8 = -1.25
A = -51.3o = 51.3o S. of E.
b. d^2 = X^2 + Y^2 = 8^2 + (-10)^2 = 164
d = 12.81 Miles.
c. d = V*t
Solve for t.
d. A = -51.3 + 180 = 128.7o, CCW = 51.3o
N. of W.
A distressed ship is 10 miles south and 8 miles east of a coast guard station.
a) What bearing should the coast guard boat take to reach the distressed ship?
b)How many miles will the coast guard boat travel?
c)If the coast guard boat is traveling 30 miles per hour, how long will it take for the boat to reach the distressed ship?
d)What bearing should the boat take to return to the coast guard station?
Please help I just came back from break and I completely forgot how to solve these kinds of problems!
1 answer
1 answer