Question

A sandstone bed has an orientation of 360/45W. If the true thickness of the bed is 50 m, the width of the outcrop exposed on a flat surface is _________ m (rounded off to one decimal place).

Hint:

Use true thickness equals outcrop width times the sine of the dip angle, then solve for the width.

Answer 1

Correct Answer: 70

Step 1: Note the numbers we have.
Strike is $360^{\circ}$, which is just north-south, dip is $45^{\circ}$ to the west, and the true bed thickness perpendicular to the layering is $t = 50$ m. We want the width of the outcrop belt as it crosses flat ground.

Step 2: Picture the cross section.
Cut a vertical section across the strike direction. The bed shows up as a slanted band tilted at $45^{\circ}$. Its true thickness is the shortest distance across that band, measured at right angles to it. The outcrop width is the horizontal distance you would walk on flat ground to cross from the top of the bed to the bottom.

Step 3: Use the right triangle formed by dip.
The true thickness, the outcrop width and the dip angle form a right triangle where
\[ t = w\sin\delta \]
so
\[ w = \frac{t}{\sin\delta} \]

Step 4: Plug in the numbers.
\[ w = \frac{50}{\sin45^{\circ}} = \frac{50}{0.7071} \]

Step 5: Work out the answer.
\[ w \approx 70.7 \text{ m} \]
This sits comfortably inside the accepted band of 70.0 to 72.0 m.
\[ \boxed{w \approx 70.7 \text{ m}} \]