But how do we figure it to accurately give equal weight to the entire line? We do it by applying basic trigonometry, finding the tangent of the angle for the known right triangle and applying it to the smaller triangle. The tangent of theta, the bearing angle, equals O. The opposite side divided by A, the adjacent side. In this case, the tangent equals 0.59 divided by 79.30, or a tangent of 0.00744.

Now, multiply the proportionate latitude, P1 by the tangent and you will get the proportionate departure, D . . .

. . . which is the proportionate position of the one-quarter section corner. In this case, 0.30 chains East but we're still not done.