Problem: Given right triangle ABC with legs 3 and 4. Find the length of the angle trisector to the hypotenuse nearer the shorter leg.
Answer Choices:
A. 13323ββ24β
B. 13123ββ9β
C. 63ββ8
D. 6510ββ
E. 1225β
Solution:
Since CD trisects right angle C,β BCD=30β and β DCA=60β and β CDE=30β, so that β³DEC is a 30ββ60ββ90β triangle. Let EC=x
β΄DE=x3β and DC=2x
x3β4βxβ=34β,x=3+43β12β,2x=3+43β24β=13323ββ24β
or
Area of β³BCD=21ββ
3β
2xβ
sin30β=23xβ
Area of β³ACD=21ββ
4β
2xβ
sin60β=23βx
Area of β³BCD+ Area of β³ACD= Area of β³ACB
β΄23βx+23βx=6,x=3+43β12β=13163ββ12β,
2x=13323ββ24β