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
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
To solve part A of this problem, you can use trigonometry. The ship is 60 miles west and 91 miles south of the harbor. To find the bearing, you need to find the angle between the ship's location and the north direction.
First, use the tangent function to find the angle:
Tangent(theta) = Opposite/Adjacent
Tangent(theta) = 60/91
Next, use the inverse tangent function to find the value of theta:
Theta = Arctan(60/91)
Theta ≈ 33.69 degrees
To round the answer to the nearest tenth of a degree, the bearing should be N33.7E.
For part B, you need to find the direct distance to the harbor. You can use trigonometry and the sine function to find the length of the hypotenuse in the triangle formed by the ship's location, the harbor, and the vertical line connecting them.
Using the sine function:
Sin(theta) = Opposite/Hypotenuse
Sin(33.69) = 60/x
To solve for x, rearrange the equation:
x * sin(33.69) = 60
Divide both sides by sin(33.69) to isolate x:
x = 60 / sin(33.69)
x ≈ 108.1667 miles
To round the answer to the nearest tenth, the direct distance to the harbor is approximately 108.2 miles.