Figure 1 below shows the cross section of a retaining wall.

Weight of Wedge ABC

From Î” ABC,

Area of Î” ABC = ½ AD Ã— BC

BC/AB = {sin(Î± + Î²)/ sin(Ï´ – Î²)}

BC = AB {sin(Î± + Î²)/ sin(Ï´ – Î²)}

Again, AD = AB sin[180^{0} – (Î± + Ï´)] = AB sin(Î± + Ï´)

**Triangle of Forces for W, P _{p} and R**

From the sine rule,

Substituting the value of W from the equation (3.34) into (3.35) we get,

In order to get the maximum value of P_{p},

Where,

Note: When Î² = 0 (leveled backfilled), Î´ = 0 (no wall friction), and Î± = 90^{0} (vertical interface between wall and backfill), then k_{a} = K_{a} = (1 – sinÉ¸)/(1 + sinÉ¸). The point of application of P_{p} is at a distance of H/3 above the base of the wall.