Question 881678
<pre>
Let M be the midpoint of BC
Let F be the foot of the perpendicular from A to BC

{{{drawing(400,2800/9,-1,8,-1,6,

triangle(0,0,7,0,6cos(.7751933733),6sin(.7751933733)),
red(line(0,0,4.937142857,3.191335208)),

locate(5.1,3.5,F),
locate(0,0,A),locate(7,0,C),locate(4.2,4.6,B),
circle(5.642857143,2.099562637,.1), locate(5.9,2.3,M)

 )}}}

We find cos(B) by using the law of cosines on triangle ABC

{{{matrix(1,3,
cos(B),""="",(a^2+c^2-b^2)/(2ac))}}}

{{{matrix(1,3,
cos(B),""="",(5^2+6^2-7^2)/(2*5*6))}}}

{{{matrix(1,3,
cos(B),""="",(25+36-49)/60)}}}

{{{matrix(1,3,
cos(B),""="",12/60)}}}

{{{matrix(1,3,
cos(B),""="",1/5)}}}

From right triangle ABF,

{{{BF/(AB)}}}{{{""=""}}}{{{cos(B)}}}

{{{BF}}}{{{""=""}}}{{{AB*cos(B)}}}

{{{BF}}}{{{""=""}}}{{{6*expr(1/5)}}}

{{{BF}}}{{{""=""}}}{{{6/5}}}

Since BC=5 and M is the midpoint of BC, and BC = 5,

BM = half of 5 or {{{5/2}}}
 
FM = BM - BF = {{{5/2 - 6/5}}} = {{{25/10-12/10}}} = {{{13/10}}} = 1.3

Edwin</pre>