The​ location, given in polar​ coordinates, of two planes are (5 mi, 14°) and (3 mi, 65°). Find the distance between the airplanes

Question

The​ location, given in polar​ coordinates, of two planes are (5 mi, 14°) and (3 mi, 65°). Find the distance between the airplanes. (Simplify your answer. Round the final answer to three decimal places as needed. Round all intermediate values to three decimal places as​ needed.)

Answer 1

Answer 2

SAS = 5-(65-14)-3

construct a triangle with 2 sides and the angle between them:

Side-Angle-Side = 5 miles- 51 degrees - 3 miles

use law of cosines, where c = side opposite C=51 degrees

c^2= a^2 +b^2 -2abcosC

c^2= 5^2 + 3^2 - 2(5)(3)cos51

c^2 =25+9-30cos51

c^2= 15.120388

c= sqr15.102388

= about 3.8885 miles apart

= about 3.889 miles

Answer 3

I'll convert from polar coordinates to cartesian coordinates and use the Pythagorean Theorem to find the distance. Along the way, I will use some trigonometric identities to make this calculation much easier.

The 2 planes are at (5 miles, 14°) and (3 miles, 65°) from the origin in the x-y plane. Converting to x-y coordinates and dropping the units, the planes are at

(5cos14, 5sin14)

and

(3cos65, 3sin65).

I will not compute sines or cosines until the very end, which will greatly simplify the calculation (not calculating until the end often helps).

Using the Pythagorean Theorem, the square of the distance between the 2 planes is

(5cos14-3cos65)² + (5sin14-3sin65)²

Performing the squares gives us

25*cos²14 - 2*5*cos14*3*cos65 + 9*cos²65 + 25*sin²14 - 2*5*sin14*3*sin65 + 9*sin²65

Rearranging terms gives us

25*cos²14 + 25*sin²14 + 9*cos²65 + 9*sin²65 - 2*5*3*cos14*cos65 - 2*5*3*sin14*sin65

Do you see where we're going? Factoring out some numbers gives us

25*(cos²14 + sin²14) + 9*(cos²65 + sin²65) - 2*3*5*(cos14*cos65 + sin14*sin65)

There are two trigonometric identities we can use here:

cos²θ + sin²θ = 1

cosθ*cosφ + sinθ*sinφ = cos(θ-φ) = cos(φ-θ)

Using these identities,

25*(1) + 9*(1) - 2*3*5*cos(65-14) =

25 + 9 - 30*cos(51) =

34 - 30*0.62932 =

15.1204

This number is the square of the distance between the 2 planes, so the distance between the two planes is

3.8885 miles.