Question

Attitudes of beds in sequences A (younger) and B (older), separated by an unconformity UU, are given in the following sectional view. If UU was horizontal when sequence A was deposited, the dip amount of beds in sequence B at that time was ...... (answer in one decimal place). 

 

Hint:

To solve dip problems, always use the law of sines or vector analysis for geometrically related bedding planes.

Answer 1

Correct Answer: 50

Present attitudes:

Sequence A: \(045^\circ, 20^\circ \text{ SE}\)

Sequence B: \(225^\circ, 30^\circ \text{ NW}\)

Note that 045° and 225° are antipodal strikes, so A and B were originally parallel beds before deformation.

Given: UU was horizontal when A was deposited
⟹ Beds of A were horizontal at that time.

Present dip of A = 20° SE, so the entire section has been tilted by 20° after deposition of A.

Present dip of B = 30° NW.
Restoring the section by untilting A by 20° increases the dip of B:

\[\text{Original dip of B} = 30^\circ + 20^\circ = 50^\circ\]