Question 1177732
<pre>

{{{drawing(400,400,-10,40,-2,20, "" ,
            line(0,2,32,2),
            line(32,2,25,9.1414),
            line(0,2,25,9.1414),
            locate(0,1.2,"A"),
            locate(32, 1.2,"B"),
            locate(25, 10.2, "C"),
            green(line(25,9.1414,25,2)),
            locate(25,1.2,"H"),
    line(25,9.1414,16,2),
    locate(16,1.2, "M"),
    locate(29,7,"a"),
    locate(10,6,"b"),
    locate(13,1.9,"c")

 
) }}}

We are given:  a=10, b=26, c=32

Using the above figure as reference, we can use the Law of Cosines to find cos(A), which is the key to solving:

{{{ cos(A) = (b^2+c^2-a^2)/(2bc) }}}


cos(A) = {{{ (26^2 + 32^2 - 10^2)/ (2*26*32) }}} = {{{ 1600/1664 }}}

( we don't need to take arccos() here, as we are going to use cos(A) )

|AH| = 26*cos(A) = 26*(1600/1664) = 25

We know |AM| = (1/2)(32) = 16 

|HM| = 25-16 = {{{ highlight( 9 ) }}} units