Limestone true thickness: A 600 m N28E, C 400 m S10W

Method: the top of the limestone is at 200 m, 220 m and 240 m depth in A, B and C. All three boreholes are vertical and the bed is 60 m thick in each (260 - 200 = 280 - 220 = 300 - 240 = 60 m), so 60 m is the vertical thickness. First find the attitude of the bed (three-point problem), then convert vertical thickness to true thickness.

Step 1: coordinates (B at origin, x east, y north)

$$\begin{aligned} A &: x = 600\sin 28^\circ = 281.68\ \text{m},\ y = 600\cos 28^\circ = 529.77\ \text{m} \\ C &: x = 400\sin 190^\circ = -69.46\ \text{m},\ y = 400\cos 190^\circ = -393.92\ \text{m} \end{aligned}$$

Step 2: plane of the top of the bed

Depth to the top: $d = 220 + a x + b y$. Substituting A and C:

$$\begin{aligned} 200 &= 220 + 281.68a + 529.77b \\ 240 &= 220 - 69.46a - 393.92b \end{aligned}$$

Solving gives $a = 0.0366$ and $b = -0.0572$ (metres of depth per metre).

Step 3: dip and dip direction

$$\tan\delta = \sqrt{a^2+b^2} = 0.0680 \Rightarrow \delta = 3.89^\circ$$

The bed becomes deeper towards the direction $(a,b)$, so the dip direction is $\arctan(a/b)$ measured from north = $147.38^\circ$ (toward SSE). The strike is perpendicular to this: $57.38^\circ$, i.e. N57.4°E.

Attitude of the bed: N57.4°E / 3.9° towards SSE (dip direction 147°).

Step 4: true thickness

$$t = t_v\cos\delta = 60\cos 3.89^\circ = 59.86\ \text{m}$$

(The bed is almost flat, so the true thickness is only slightly less than 60 m.)

Answer: true thickness of the limestone = 59.86 m (about 59.9 m); bed dips 3.9° towards SSE.