Reserve: quartzite apparent dips 1:9 and 1:12, 2.7 x 3.1 km

Step 1: true dip from the two apparent dips

For a direction of bearing $\beta$ the gradient of the bed satisfies $g_E\sin\beta + g_N\cos\beta = \tan\delta_a$ ($g_E$, $g_N$ = east and north components of the true gradient).

$$\begin{aligned} g_E\sin 20^\circ + g_N\cos 20^\circ &= \tfrac{1}{9} = 0.11111 \\ g_E\sin 65^\circ + g_N\cos 65^\circ &= \tfrac{1}{12} = 0.08333 \end{aligned}$$

Solving: $g_E = 0.04434$, $g_N = 0.10211$.

$$\tan\delta = \sqrt{0.04434^2 + 0.10211^2} = 0.1113 \Rightarrow \delta = 6.35^\circ\ (1:8.98)$$

Dip direction $= \arctan(0.04434/0.10211) = 23.5^\circ$ (N23.5°E).

Step 2: true thickness

$$t = t_v\cos\delta = 105\cos 6.35^\circ = 104.36\ \text{m}$$

Step 3: volume (reserve)

Area of the terrain $= 2.7\ \text{km} \times 3.1\ \text{km} = 8.37\ \text{km}^2 = 8.37\times 10^6\ \text{m}^2$

$$V = A \times t = 8.37\times 10^6 \times 104.36 = 8.73 \times 10^8\ \text{m}^3$$

Answer: true dip = 6.35° (1:9.0) towards N23.5°E; true thickness = 104.4 m; reserve of quartzite = 8.73 × 10⁸ m³ (0.873 km³).