Limestone reserve for cement factory

Step 1: true dip from the two apparent dips

Apparent dips: 1:16 along S15°E (bearing 165°) and 1:22 along S27°W (bearing 207°).

$$\begin{aligned} g_E\sin 165^\circ + g_N\cos 165^\circ &= \tfrac{1}{16} = 0.06250 \\ g_E\sin 207^\circ + g_N\cos 207^\circ &= \tfrac{1}{22} = 0.04545 \end{aligned}$$

Solving: $g_E = 0.01761$, $g_N = -0.05999$.

$$\tan\delta = \sqrt{0.01761^2 + 0.05999^2} = 0.06252 \Rightarrow \delta = 3.58^\circ\ (1:16.0)$$

Dip direction $= 163.6^\circ$ (S16.4°E).

Step 2: true thickness

$$t = 215\cos 3.58^\circ = 214.58\ \text{m}$$

Step 3: volume

The extent of the rock is read as 3.27 km $\times$ 72.05 m (length $\times$ width in plan):

$$A = 3270 \times 72.05 = 2.356\times 10^5\ \text{m}^2$$ $$V = A \times t = 2.356\times 10^5 \times 214.58 = 5.06 \times 10^7\ \text{m}^3$$

Step 4: mass (reserve in tonnes)

$$\text{Mass} = V \times \rho = 5.056\times 10^7 \times 2.75\times 10^3 = 1.39 \times 10^{11}\ \text{kg} = 1.39 \times 10^{8}\ \text{tonnes}$$

Answer: true dip = 3.58° (1:16) towards S16.4°E; true thickness = 214.6 m; volume = 5.06 × 10⁷ m³; reserve of limestone = 1.39 × 10¹¹ kg (about 139 million tonnes).