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.

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

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

$\displaystyle c^2 = 17^2 + 20^2 - 2ac(cos 110^o)$

$\displaystyle c^2 = 289 + 400 - 680(cos 110^0)$

$\displaystyle c^2 = 679.33$

$\displaystyle {\sqrt{c^2}} = {\sqrt{679.33}}$

$\displaystyle c = 21.06$

So the answer should be 21.06 km. But, according to the answer key, it's 30.36 km. I cannot figure out where I erred. Could someone please help?