One quick question (half of answer provided)?

Question

An airplane flies 200 km due west from city A to city B and then 255 km in the direction of 34.0° north of west from city B to city C. In straight-line distance, it is 437.7 km from city C from city A

Relative to city A, in what direction is city C?

in degrees

in degrees north of west

Answer 1

the angle the plane would have to fly to get from A to C is 19.12 degrees north of west.

first find how far north C is from B (which is also how far north C is from A)..this distance is 255 sin 34

now find how far west C is from B...this distance is 255 cos 34

so...the angle from A to C is going to be tan of the north distance/ total west distance....

tan θ=[(255sin 34)/(200+255cos34)]=19.12

Answer 2

How did you figure out the straight-line distance? Use the North and west components of displacement to find the direction of travel.

Answer 3

So ... as a manner to locate the right this moment-line distance you probable had to be sure the area traveled regularly occurring west, and the area traveled regularly occurring north, and create a terrific triangle, with the right this moment-line with the aid of fact the hypotenuse and a nook at city A. using that individual same triangle, that's common to apply inverse tangent or one in all the different trig applications to be sure the perspective at nook A, which would be process shuttle relative to due west.

Answer 4

Eleventy four

Answer 5

WNW