+0  
 
-2
48
3
avatar+227 

Given triangle ABC, in which angle ACB is 90 degrees.

 

CE and CD trisect angle ACB, AC=5, BC=12

Prove that CE = 40(12-5sqrt3)/23

 Apr 8, 2020
 #1
avatar+930 
0

Deleted...

 Apr 8, 2020
edited by HELPMEEEEEEEEEEEEE  Apr 8, 2020
 #2
avatar
+1

cos(B) = 12/13,   sin(B) = 5/13,   angle CEB = 180 - 30 - B = 150 - B.

Using the sine rule in triangle CEB,

12/sin(150 - B) = CE/sin(B),

so

CE = 12sin(B)/ sin(150 - B) = 12sin(B)/(sin150cos(B) - cos150sin(B)).

Substitute for sin(B), cos(B), sin and cos 150 and simplify.

You get the stated result.

(Note regarding the previous answer. Angle A does not equal 60 degrees.)

 Apr 8, 2020
 #3
avatar+930 
0

That is definitely a better answer thanks!

 Apr 8, 2020

14 Online Users

avatar
avatar