An airplane flies 200 km due west from city A to city B and then 240 km in the direction of 31.0° north of west from city B to city C.

(a) In straight-line distance, how far is city C from city A?

(b) Relative to city A, in what direction is city C?

Law of cosines

b^2 = a^2+c^2-2ac cos B
b^2 = 240^2 + 200^2 -2(240)(200)cos(180-31)

solve for b which is distance from A to C

then law of sines
sin A/240 = sin(180-31)/b
solve for angle A

As a sailboat sails 59.0 m due north, a breeze exerts a constant force 1 on the boat's sails. This force is directed at an angle west of due north. A force 2 of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just 36.0 m. What is the angle between the direction of the force 1 and due north?

because it says the two forces do the same amount of work, you can tell that the two distances given are the hypotenuse and leg of a right triangle. after that it's just a matter of taking the inverse cos of adjacent/hypotenuse

To find the straight-line distance between city A and city C, we can use the concept of vector addition.

(a) To begin, we can represent the first leg of the airplane's journey from city A to city B as a vector going due west with a magnitude of 200 km. We'll call this vector AB.

Next, we can represent the second leg of the journey from city B to city C as a vector going in the direction of 31.0° north of west with a magnitude of 240 km. We'll call this vector BC.

To solve for the straight-line distance between city A and city C, we need to find the sum of vectors AB and BC. This can be done by breaking down each vector into its horizontal (x-component) and vertical (y-component) parts.

For vector AB:
Magnitude = 200 km
Horizontal component = 200 km (due west)
Vertical component = 0 km (since it's going straight west)

For vector BC:
Magnitude = 240 km
Horizontal component = 240 km · cos(31.0°) (westward component)
Vertical component = 240 km · sin(31.0°) (northward component)

Now, we can add up the horizontal and vertical components separately for each vector:

Horizontal component of AC = Horizontal component of AB + Horizontal component of BC
Vertical component of AC = Vertical component of AB + Vertical component of BC

Finally, we can use the Pythagorean theorem to find the magnitude of vector AC:

Magnitude of AC = √[(Horizontal component of AC)^2 + (Vertical component of AC)^2]

This will give us the straight-line distance between city A and city C.

(b) To find the direction of city C relative to city A, we can use trigonometry.

The direction is usually given as an angle measured from the positive x-axis. We can find this angle by taking the inverse tangent (or arctan) of the ratio of the vertical component to the horizontal component:

Angle θ = tan^(-1)((Vertical component of AC) / (Horizontal component of AC))

This angle will give us the direction of city C relative to city A.