prove that the area of any triangle XYZ is y^2sin(x)sin(z)/2*sin(y)

1 answer

y^2sin(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

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prove that the area of any triangle XYZ is y^2*sin(x)*sin(z)/2*sin(y)

y^2*sin(x)*sin(z)/2*sin(y)

prove that the area of any triangle XYZ is y^2sin(x)sin(z)/2*sin(y)

y^2sin(x)sin(z)/2*sin(y)