1) in triangle ABC, AD is the angle bisector of < A. If BD = 5, CD = 6 and AB = 8, find AC.

Answer 1

Angle at C is common to triangles
ACD and to ACB
since angles in both triangles add to 180, angle ADC = twice angle ABD
call angle CAD = T
then BAD = T
call angle ABD = A
then angle ADC = 2A

then
sin T / 6 = sin 2A /x
sin 2T / 11 = sin A /x
sin T / 5 = sin (180-2A)/8 = sin 2A / 5

sin 2T = 2 sin T cos T
sin 2A = 2 sin A cos A

sin T/6 = 2 sin A cos A/x
2 sin T cos T/11 = sin A /x
sin T /5 = 2 sin A cos A /5

sin T = 12 sin A cos A /x
sin T = 11 sin A /(2x cos T)
so
12 cos A = 11/ (2 cos T)
cos A cos T = 11/24
and
sin T = 2 sin A cos A
but
sin T = 12 sin A cos A/x
so
12/ x = 2
x = 6
I bet there are easier ways :)

Answer 2

Thanks

Answer 3

No problem. You are welcome.