Question 1148362
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;There is &nbsp;&nbsp;<U>EXTREMELY &nbsp;SIMPLE</U> &nbsp;solution.



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From the diagram,  we have an isosceles triangle AXO with equal sides lengths  | AX | = | OX |.


The base AO of this triangle is 16 units long, since it is the hypotenuse of the (30° - 60° - 90°) triangle AOT.


The angle A is 30°.


Draw the altitude  XZ  in the triangle AXO.

Then you will get right angled triangle AXZ with the long leg  AZ of the length of 8 = 16/2 units and the angle A of 30°.


Thus  cos(A) = cos(30°) = {{{8/m}}} = {{{sqrt(3)/2}}},


which implies  m = {{{(2*8)/sqrt(3)}}} = {{{16/sqrt(3)}}} = 9.237 units approximately, if you want to have the numerical value.


<U>ANSWER</U>.  m = {{{16/sqrt(3)}}} = 9.237 (approximately).
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Solved.