The center of mass (the same with the center of weight) of a right triangle is 1/3 distance from the right angle on both legs.

The momentum of the triangle weight plus that of the force in B with respect to point A is zero (the plate does not rotate about A)

`G*(l/3) -F_B*l =0`

`F_B =G/3` directed upwards.

Now for static equilibrium one needs that on the vertical axis the sum of all forces is zero.

`F_A + F_B -G = 0`

`F_A =G-F_B =G-G/3 =(2/3)*G =(2/3)*m*g`

**The correct answer is B)** `(2*mg)/3`

**Further Reading**

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