1. Let A represent the initial position of the plane and C the runaway (check the picture at the bottom of the answer).
2. The angle shown in the circle and angle C are alternate internior angles, or Z angles, so the measure of angle C is 15°.
3. [tex]tan 15^o= \frac{2530}{BC} [/tex]
4. Angle 15° is generally handled with half angle formulas, we have
[tex]tan( \frac{x}{2}) = \sqrt{ \frac{1-cosx}{1+cosx}} [/tex]
so
[tex]tan( \frac{30}{2}) = \sqrt{ \frac{1-cos30}{1+cos30}}= \sqrt{ \frac{1- \sqrt{3}/2 }{1+ \sqrt{3}/2}} = \sqrt{ \frac{2- \sqrt{3} }{2+ \sqrt{3} }} = \sqrt{ \frac{0.268}{3.732}}= \sqrt{0.072} =0.27 [/tex]
5. Another way is to use a trigonometric values calculating software.
6. Thus, [tex]BC=2530/tan15^o=2530/0.27=9370 (ft)[/tex]
