# Thread: The distance between two planes

1. ## The distance between two planes

The Problem: One plane is 20 km from the Winnipeg Airport. A larger plane is 17 km from the airport. If the angle between the sightings is 110 degrees, how far apart are the planes?

Attempted Solution: First, I drew a triangle.

Then I used the law of cosines.

$c^2 = a^2 + b^2 -2ac(cos \theta)$

Then put in the $a$, $c$, and $cos \theta)$ values.

$c^2 = 17^2 + 20^2 - 2ac(cos 110^o)$
$c^2 = 289 + 400 - 680(cos 110^0)$
$c^2 = 679.33$
${\sqrt{c^2}} = {\sqrt{679.33}}$
$c = 21.06$

2. The part where you went wrong is that the Law of Cosine you used isn't correct.

$c^2 = 17^2 + 20^2 - 2{\color{red}ac}(cos 110^o)$
It should be ab.

Let A, B, and C be the vertices of a triangle and a, b, and c be the sides of the same triangle. Side a is opposite vertex A, side b is opposite vertex B, side c is opposite vertex C.

The Law of Cosines state that:
$a^2 = b^2 + c^2 - 2bc\cos{A}$

$b^2 = a^2 + c^2 - 2ac\cos{B}$

$c^2 = a^2 + b^2 - 2ab\cos{C}$

3. Originally Posted by D. Martin
The Problem: One plane is 20 km from the Winnipeg Airport. A larger plane is 17 km from the airport. If the angle between the sightings is 110 degrees, how far apart are the planes?

Attempted Solution: First, I drew a triangle.

Then I used the law of cosines.

$c^2 = a^2 + b^2 -2ac(cos \theta)$

Then put in the $a$, $c$, and $cos \theta)$ values.

$c^2 = 17^2 + 20^2 - 2ac(cos 110^o)$
$c^2 = 289 + 400 - 680(cos 110^0)$
$c^2 = 679.33$
${\sqrt{c^2}} = {\sqrt{679.33}}$
$c = 21.06$