Gradient A to B, RL of P = 1234.560

Staff is vertical, $k = 100$, $c = 0$, angles are zenith angles $z$, so the vertical angle is $\theta = 90^\circ - z$.

$$H = k\,s\sin^2 z, \qquad V = \frac{k\,s}{2}\sin 2\theta$$

From station A

• $s = 2.025 - 1.025 = 1.000$ m, $z = 75^\circ 35' 25''$, $\theta = +14°24'35"$ (elevation)
• $H_A = 100\times 1.000\times\sin^2 z = 93.807$ m
• $V_A = 50\times 1.000\times\sin 2\theta = +24.103$ m
• RL of P = RL of A + HI + V − r, so $RL_A = 1234.560 - 24.103 + 1.525 - 1.42 = 1210.562$ m

From station B

• $s = 2.125 - 0.925 = 1.200$ m, $z = 96^\circ 52' 45''$, $\theta = -6°52'45"$ (depression, so $V$ is negative)
• $H_B = 100\times 1.200\times\sin^2 z = 118.278$ m
• $V_B = 50\times 1.200\times\sin 2\theta = -14.270$ m
• $RL_B = 1234.560 - (-14.270) + 1.525 - 1.48 = 1248.875$ m

Horizontal distance AB

Bearing of AP = 60°, bearing of BP = 345°, so the bearing of PA = 240° and of PB = 165°. The angle at P is $240^\circ - 165^\circ = 75^\circ$.

            P
           /|\
    H_A   / | \  H_B
         /  75°\
        A ------ B

$$AB = \sqrt{H_A^2 + H_B^2 - 2H_AH_B\cos 75^\circ} = \sqrt{93.807^2 + 118.278^2 - 2(93.807)(118.278)\cos 75^\circ} = 130.561\ \text{m}$$

Gradient

Difference in RL = $RL_B - RL_A = 1248.875 - 1210.562 = 38.312$ m

$$\text{Gradient} = \frac{38.312}{130.561} = 0.2934 = 1\text{ in }3.41$$

Answer: the ground rises from A to B at a gradient of 1 in 3.41 (about 29.3 %); AB = 130.56 m and B is 38.31 m higher than A.