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: Read the given orientation.
The bed strikes at \(360^{\circ}\), which is the same as \(000^{\circ}\), meaning it runs north to south, and it dips \(45^{\circ}\) towards the west. The true thickness of the bed, measured perpendicular to the bedding plane, is given as \(t = 50\) m.

Step 2: Set up the geometry on flat ground.
On a horizontal, flat ground surface, a line drawn across strike, from the top contact of the bed to the bottom contact, is the outcrop width, call it \(w\). This width, the bed's dip angle \(\delta\), and the true thickness \(t\) form a right triangle, because the true thickness is measured perpendicular to the dipping bed while the outcrop width is measured along the horizontal ground surface.

Step 3: Write the relation between true thickness and outcrop width.
For a bed of dip \(\delta\) cropping out on flat ground, the true thickness relates to the outcrop width as
\[ t = w \sin\delta \]
Rearranging for the outcrop width,
\[ w = \frac{t}{\sin\delta} \]

Step 4: Substitute the given values.
\[ w = \frac{50}{\sin 45^{\circ}} \]
Since \(\sin 45^{\circ} = \dfrac{1}{\sqrt{2}} \approx 0.7071\),
\[ w = \frac{50}{0.7071} \]

Step 5: Compute the final value.
\[ w \approx 70.7 \text{ m} \]

Step 6: Final conclusion.
The width of the outcrop exposed on the flat surface is about 70.7 m, which lies in the accepted range of 70.0 to 72.0 m.
\[ \boxed{w \approx 70.7 \text{ m}} \]