Another problem I need to confirm:
A ship is 60 miles west and 91 miles south of the harbor.
a) What bearing should the ship take to sail directly to the harbor? (Round answer to nearest tenth of degree.)
b) What is the direct distance to the harbor?
A)
Tan(theta) = 60/91
Theta = Arctan(60/91)
Theta = 33.69
Answer: N33.7E
B)
Sin(theta) = Opposite/Hypotenuse
Sin(33.69)=60/x
xsin(33.69)=60
x=60/sin(33.69(
x=108.1667 miles
Answer: x = 108.2 miles
2 answers
sweet correct
tanØ = 60/91
Ø = arctan(60/91) = 33.398°
distance , using Pythagoras:
d^2 = 60^2 + 91^2 = 11881
d= √11881
= 109 miles
Ø = arctan(60/91) = 33.398°
distance , using Pythagoras:
d^2 = 60^2 + 91^2 = 11881
d= √11881
= 109 miles
1 answer