+236 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
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.)
