Question 1148503
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1.  Make a sketch.



2.  Due to symmetry,  angle BMP = angle AMQ = 60°.



3.  Consider triangle BMP.

    Its side BM is 36 cm;  its side MP is unknown; let it be "a" cm long.

    Its angle BMP is 60°; its angle MBP is 45°.

    Its angle MPB = 180° - 60° - 45° = 105°.



4.  Apply the sine law theorem to triangle BMP.


        {{{36/sin(105^o))}}} = {{{a/sin(45^o)}}}.    (1)


    Use  sin(105°) = sin(180°-105°) = sin(75°) = sin(45°+30°) = sin(45°)*cos*30°) + cos(45°)*sin(30°) = {{{(sqrt(2)/2)*(sqrt(3)/2) + (sqrt(2)/2)*(1/2)}}} = {{{(sqrt(6)+sqrt(2))/4}}}.


    Use  sin(45°) = {{{sqrt(2)/2}}}.


    Then from (1)  you get


        a = {{{(36*sin(45^o))/sin(105^o)}}} = {{{36*((sqrt(2)/2)/((sqrt(6)+sqrt(2))/4))}}} = {{{36*((2*sqrt(2))/(sqrt(6)+sqrt(2)))}}}.   (2)


    It is the final formula for the unknown side length "a".


    If in addition to the final formula (2) you need its numerical value,

    here it is   a = 26.354 centimeters,  with 3 correct decimal places after the decimal dot.


     Now, if you want to find the area of the triangle MPQ, use the formula


        area = {{{(a^2*sqrt(3))/4}}} = {{{(26.354^2*sqrt(3))/4}}} = 300.74 cm^2.
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Solved.