Try drawing the triangle
Angle # A=(5pi)/12# and angle #C=pi/4# and side #b=4#
side #b# is the base of the triangle. There 's a need to solve for the height #h# from angle B to side b to compute the area.
#h cot A+h cot C=b#
#h=b/(cot A+cot C)=4/(cot ((5pi)/12)+cot (pi/4)#
From double angle formulas:
#cot ((5pi)/12)=cos(pi/4+pi/6)/sin(pi/4+pi/6)=(cos (pi/4)*cos (pi/6)-sin (pi/4)*sin (pi/6))/(sin (pi/4)*cos (pi/6)+cos (pi/4)*sin (pi/6))#
#cot ((5pi)/12)=(sqrt3-1)/(sqrt3+1)#
Solve #h# now
#h=4/(cot ((5pi)/12)+cot (pi/4))=4/((sqrt3-1)/(sqrt3+1)+1)#
#h=2/3*(3+sqrt3)#
Solve #Area=1/2*b*h#
#Area=1/2*4*2/3*(3+sqrt3)#
#Area=4/3*(3+sqrt(3))#
#Area=6.3094# square units
Have a nice day !!! from the Philippines .
