Question 43248
In triangle ABC, BD is the median which bisects the side AC. 
Drop a perpendicular from C on AB intersecting AB at E.


{{{drawing(400,200,-10,10,-2,8,
line(-5,0,5,0),
line(-5,0,0,4.196),
line(0,4.196,5,0),
line(5,0,-2.5,2.098),
line(0,4.196,0,0),
locate(5.5,0,B),
locate(-5.5,0,A),
locate(0,5,C),
locate(-3.5,3,D),
locate(0,-0.5,E)
)}}}


In triangle ABC, < CAB = < CBA = {{{40^o}}} and AB = 10 units.
As triangle ABC is isoscles, so the perpendicular dropped from the vertex on the opposite bisects the opposite side.
Hence, AE = BE = {{{AB/2}}} = {{{10/2}}} = 5 units.


Now, in triangle AEC, < AEC is right angle.
So, {{{AE/AC}}} = cos(< CAE)
or {{{5/AC = cos(40^o)}}}
or {{{AC = 5/cos(40^o)}}}
or {{{AC = 5/0.839}}}
or {{{AC = 6.527}}}


So, AD = {{{1/2}}}AC = {{{6.527/2}}} = 3.263 units.


Now, consider triangle ADB.
Apply cosine formula
{{{BD^2 = AD^2 + AB^2 - 2*AD*AB}}}cos(< BAD)
or {{{BD^2 = 3.263^2 + 10^2 - 2*3.263*10*cos(40^o)}}}
or {{{BD^2 = 60.655}}}
or {{{BD = sqrt(60.655)}}}
or BD = 7.788 units


Thus, the length of the median to the line segment AC is 7.788 units.