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# Distance between two objects?

The locations, given in polar coordinates, of the two planes approaching the Ontario airport are (7 mi, 48°) and (5 mi, 125°). Find the distance between the two points.

### 3 Answers

- az_lenderLv 71 month ago
Wayne has an arithmetic error.

49 and 25 is 74, not 64.

Anyway, it's easiest to get the 77-degree angle by just subtracting 48 from 125.

Then

d^2 = 49 + 25 - 2*5*7*cos(77)

= 74 - 70*cos(77) = 58.253

=> d = 7.6 miles.

- PopeLv 71 month ago
The angle between the two vectors is 125° - 48° = 77°. What you have now is two sides of a triangle and the included angle. You want the third side. Use the cosine rule.

√[7² + 5² - 2(7)(5)cos(77°)] ≈ 7.63 mi

- Wayne DeguManLv 71 month ago
The plane at (7 mi, 48°) makes an angle of 48° with the easterly direction.

The plane at (5 mi, 125°) makes an angle of 55° with the westerly direction

Hence, angle between the directions is 180° - (48° + 55°)

i.e. 77°

Using the 'cosine rule' we get:

d² = 7² + 5² - 2(7)(5)cos77°

so, d² = 64 - 70cos77°

Hence, d² = 48.3

Therefore, d = 6.9 miles

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