Word Problem Please Help

Question

An airplane flies 200 km due west from city A to city B and then 300 km in the direction of 30.0∘ north of west from city B to city C. How far is city C from city A?

Answer 1

when we draw the figure, the path traveled by a to b then to cmakes to sides of a triangle... lets name them...

from city a to b, we name it side A

from city b to c, we name it side B

from city a to c(what we are looking for), we name itC...

lets name the angle between side C and A φ...

now the angle between A and B is 150^o (we name thisθ)which is the suplementary angle to30^o(given). (draw it and you'll understand)

 

 (a) using cosine law we can find C

 

C²=A² + B² - 2ABcosθ,since θ is the opposite angle to side C..

 

C²=(200)²+(300)² -2(200)(300)cos(150^o)

C²=233923.0485km²

C=483.66km

 

(b) now using sine law,

 

 

 

 

φ=18.067^o

 

therefore relative to city a, city c is18.067^o NW

Answer 2

draw a diagram and use the law of cosines

call "x" the distance from A to C

x^2=200^2+300^2-2*200*300*cos150°

x^2=40,000+90,000-120,000*(-0.8660254)

x^2=130,000+103,903

x^2=233,923

x=√233,923

x=483.6558 miles (round as needed)

Answer 3

Let city A be at location (0, 0), west be along the negative x axis, and north along the positive y axis.

City B is then located at (-200, 0}.

From city B we move west by 300*cos(30) and north by 300*sin(30) putting city C at (-200 - 259.8, 0 + 150} or (-459.8, 150)

This puts city C at a distance of sqrt{459.8^2 + 150^2) = 483.6 km from city A.