tan15^o

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^o. 

4.        So angle CDE=15^o   (since angle AED=(150^O- 30^o)/2=75^o). 

5.       Since tan 15^o =CE/CD        tan 15^o  =(2- sqrt(3))/1.  
 

6.       finally  tan 15^o = 2- sqrt(3) 

7.       The above equals      approximately   .268

2nd part later (time allowing)

Edited on January 3, 2014, 2:48 pm

Posted by Ady TZIDON on 2014-01-03 14:45:39