tan15°
1. Draw a rectangle ABCD s.t AB=CD=1 ;AD=BC=2.
2. On the side CB mark a point E s.t. AD=AE= 2 .
3. Now CE=2- sqrt(3) angle BAC=60°
4. So angle CDE=15° (since angle AED=
(150O- 30o)/2=75°).
5. Since tan 15° =CE/CD tan 15° =(2- sqrt(3))/1.
6. finally tan 15° = 2- sqrt(3)
7. The above equals approximately .268
1st part later (time allowing)
Ady Tzidon
Edited on January 5, 2014, 12:39 pm