y^2*sin(x)*sin(z)/2*sin(y)
area = z/2 * y sinX
but, z/sinZ = y/sinY, so
z = ysinZ/sinY
area = ysinZ/2sinY * y sinX
= y^2 sinZ sinX / 2sinY
prove that the area of any triangle XYZ is y^2*sin(x)*sin(z)/2*sin(y)
1 answer