Question 1148427
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The altitude of the triangle ABC from vertex A, drawn to the side BC  is  

    h = {{{22*(sqrt(3)/2)}}} cm.


Let the center of the circle be the point O at this altitude.


And let the points D and E are the tangent points at the sides AB and AC.


Then the triangle ODA is a right angled triangle with the shorter leg of 5 cm and the acute angle OAD of 30 degrees.


Hence, its hypotenuse OA is twice the radius. i.e. 10 cm.


Thus the distance from the center O of the circle to the side BC is

    d = {{{22*(sqrt(3)/2)}}} - 10 cm = 19.053 - 10 = 9.053 cm.


<U>ANSWER</U>.  The distance from the center O of the circle to the side BC is  d = {{{22*(sqrt(3)/2)}}} - 10 cm = 9.053 cm.
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Solved.