Method using the double angle tangent formula: Drop a perpendicular from A to BC and label the intersection as point F. Drop a perpendicular from E to BC and label the intersection as point G. Consider ΔCEG. It is a 30°-60°-90° special right triangle, with hypotenuse of length 4, so CG = 2 (opposite the 30° angle) and EG = 2√3. Consider ΔCAF. It is also a 30°-60°-90° special right triangle, but with hypotenuse of length 8, so CF = 4 (opposite the 30° angle) and AF = 4√3. tan(
Method using the double angle tangent formula: Drop a perpendicular from A to BC and label the intersection as point F. Drop a perpendicular from E to BC and label the intersection as point G. Consider ΔCEG. It is a 30°-60°-90° special right triangle, with hypotenuse of length 4, so CG = 2 (opposite the 30° angle) and EG = 2√3. Consider ΔCAF. It is also a 30°-60°-90° special right triangle, but with hypotenuse of length 8, so CF = 4 (opposite the 30° angle) and AF = 4√3. tan(
I have an alternative solution with trigonometry to share with.
Denoting alpha as a