Try drawing the triangle
Angle A=(5pi)/12A=5π12 and angle C=pi/4C=π4 and side b=4b=4
side bb is the base of the triangle. There 's a need to solve for the height hh from angle B to side b to compute the area.
h cot A+h cot C=bhcotA+hcotC=b
h=b/(cot A+cot C)=4/(cot ((5pi)/12)+cot (pi/4)h=bcotA+cotC=4cot(5π12)+cot(π4)
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(5π12)=cos(π4+π6)sin(π4+π6)=cos(π4)⋅cos(π6)−sin(π4)⋅sin(π6)sin(π4)⋅cos(π6)+cos(π4)⋅sin(π6)
cot ((5pi)/12)=(sqrt3-1)/(sqrt3+1)cot(5π12)=√3−1√3+1
Solve hh now
h=4/(cot ((5pi)/12)+cot (pi/4))=4/((sqrt3-1)/(sqrt3+1)+1)h=4cot(5π12)+cot(π4)=4√3−1√3+1+1
h=2/3*(3+sqrt3)h=23⋅(3+√3)
Solve Area=1/2*b*hArea=12⋅b⋅h
Area=1/2*4*2/3*(3+sqrt3)Area=12⋅4⋅23⋅(3+√3)
Area=4/3*(3+sqrt(3))Area=43⋅(3+√3)
Area=6.3094Area=6.3094 square units
Have a nice day !!! from the Philippines .
