3-Point Problem and Tectonic Plates: 04-BS-14, December 2016, Q3

Question 3: 3-Point Problem and Tectonic Plates (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

23. Strike, dip and outcrop trace of the coal seam

Given. The coal seam crops out at point A. True dip is 14° toward the West (bearing 270°). North is up and the scale bar reads 100 m. The labelled contours are 100, 150, 200 and 250 m, with one unlabelled contour between each pair, so the contour interval is 25 m. Counting inward from the 100 m contour (100, 125, 150, 175, 200, 225, 250, 275 m summit ring), A lies on the fourth contour, so A is at 175 m.

Find. (a) Strike and dip. (b) The outcrop trace of the seam on the contour map. (c) The apparent dip in the direction S60°W.

Approach. Strike is horizontal and perpendicular to the true-dip direction. The outcrop is found with structure contours: lines of equal elevation on the seam, drawn parallel to strike and spaced by the contour interval divided by tan(dip). The seam crops out wherever a structure contour crosses the topographic contour of the same elevation. The apparent dip uses the supplied formula, where θ is the map angle between the true-dip direction and the direction of interest.

(a) Strike and dip. The dip direction is due West (270°). Strike is perpendicular to it: $270^{\circ}\pm 90^{\circ}$ = 180° and 360°, a North–South line. $$\boxed{\text{Strike N–S (000}^{\circ}\text{/180}^{\circ}\text{)}, \quad \text{Dip } 14^{\circ}\text{ W}}$$ In quadrant notation this is N00°E (N–S), 14°W. In azimuth (right-hand-rule) notation it is 180/14.
(b) Structure-contour spacing. For one 25 m contour interval: $$d = \frac{25\text{ m}}{\tan 14^{\circ}} = \frac{25}{0.2493} \approx \boxed{100.3\text{ m}}$$ This is almost exactly the length of the map's 100 m scale bar. Because the seam dips West, its elevation falls 25 m for every 100.3 m westward and rises 25 m for every 100.3 m eastward. So draw N–S structure contours at: 175 m through A; 150 m 100.3 m west of A; 125 m 200.6 m west; 100 m 300.9 m west; and 200 m 100.3 m east.

(b) Plotting the trace. Mark every point where a structure contour crosses the topographic contour of the same value, then join the points. The positions below were read from the printed map (scale calibrated from the scale bar, ±5 m):

Structure contour	Position	Meets topographic contour	Outcrop points (relative to A)
175 m	through A	175 m (the contour A sits on)	A itself: the seam's outcrop reaches its easternmost point here
150 m	100 m W of A	150 m	about 3 m N (in the saddle) and 63 m S of A
125 m	200 m W of A	125 m	about 36 m N and 83 m S of A
100 m	301 m W of A	100 m	about 31 m N and 83 m S of A
200 m	100 m E of A	none. This line lies east of the whole hill, where the ground is below 100 m.	no outcrop east of A

Between two points the trace runs where the ground elevation equals the seam elevation. For example, 50 m west of A the seam is at $175 - 50\tan 14^{\circ} \approx 162.5$ m, so the trace lies about halfway between the 150 and 175 m topographic contours. Each branch therefore stays between the two topographic contours that bracket its seam elevation. The result is a V (U) shape pointing east with its tip at A. Its north branch passes north of the eastern knob, drops into the saddle and runs west between the 125 and 150 m contours. Its south branch runs west along the southern slope. Both branches leave the mapped contours on the west side, below 100 m. Under the hill (including both summits) the seam is buried, because the ground is higher than the seam. East of the trace the ground is lower than the seam, so the seam has been eroded away.

(c) Apparent dip toward S60°W. S60°W is bearing $180^{\circ}+60^{\circ}=240^{\circ}$. The angle between the true-dip direction (270°) and this direction is $\theta = |270^{\circ}-240^{\circ}| = 30^{\circ}$. With $\alpha = 14^{\circ}$: $$\alpha' = \arctan(\cos 30^{\circ}\times\tan 14^{\circ}) = \arctan(0.8660\times0.2493) = \arctan(0.2159)$$ $$\boxed{\alpha' \approx 12.2^{\circ}\ \text{toward S60°W}}$$ As expected for any direction other than the true-dip direction, $\alpha' < \alpha$ (12.2° < 14°).

[Figure not reproduced: Figure Q23 with structure contours and traced outcrop. See the official exam paper or the cited reference text.]

Figure Q23 with the N–S structure contours (dashed blue, 100.3 m apart) and the traced coal-seam outcrop (red). The circles mark where each structure contour meets the topographic contour of the same elevation. The trace across the north-east lobe is interpolated between the 150 and 175 m contours, which the printed map leaves partly unclosed there.

24. Tectonic plates and boundaries (Figure Q24)

[Figure not reproduced. See the official exam paper.]

Figure Q24 as printed. The map is Pacific-centred: Europe and Africa are at the left, Asia and Australia are left of centre, and the Americas are at the right.

Locate the map by its continents first. Label 1 sits over Europe and northern Asia. Label 2 sits in the Indian Ocean, west of Australia, within the plate that carries India and Australia. Label 3 sits in the middle of the largest ocean, the Pacific. Label A sits on the boundary along the northern rim of the Pacific (the Aleutian arc, between Alaska and Kamchatka). Label B sits on the boundary running down the middle of the South Atlantic, halfway between South America and Africa.

Label	Answer	Reasoning
Plate 1	Eurasian Plate	Carries Europe and most of Asia (upper left of the map).
Plate 2	Indo-Australian Plate	Carries India, Australia and the eastern Indian Ocean (on some maps it is split into the Indian and Australian plates).
Plate 3	Pacific Plate	The largest plate, almost entirely oceanic, filling the centre of this Pacific-centred map.
Boundary A	Convergent (subduction zone)	The Aleutian Trench. Here the Pacific Plate subducts northward beneath the North American Plate, producing the Aleutian volcanic island arc and deep earthquakes.
Boundary B	Divergent (mid-ocean ridge, spreading centre)	The Mid-Atlantic Ridge between the South American and African plates. New ocean floor forms here as the two plates move apart, so the ridge runs parallel to both coastlines.

Item	Result
23(a)	Strike N–S (000°/180°), dip 14° W
23(b)	Contour interval 25 m; A at 175 m; structure contours N–S every 100.3 m; eastward-pointing V-shaped outcrop with its tip at A
23(c)	Apparent dip 12.2° toward S60°W
24(a)	1 = Eurasian, 2 = Indo-Australian, 3 = Pacific
24(b)	A = convergent (Aleutian subduction zone); B = divergent (Mid-Atlantic Ridge)

Topic: Structural geology (strike/dip, apparent dip, structure contours, three-point problem) and plate-tectonic boundary classification.

Key relations: $\alpha' = \arctan(\cos\theta \tan\alpha)$ (apparent dip); structure-contour spacing $d = CI/\tan\alpha$; strike $\perp$ dip direction; divergent (spreading ridge) vs. convergent (subduction, trench/arc) vs. transform boundary map signatures.

Why this works: A structure contour is simply the trace of the dipping plane at one elevation — treating the coal seam exactly like a tilted topographic surface lets the same contour-intersection logic used for reading a topo map locate the outcrop; the apparent-dip formula is the projection of the true dip vector onto a vertical plane rotated by $\theta$ from the true-dip direction.

Common pitfall: Forgetting that apparent dip is always $\le$ true dip (a common arithmetic-sign error inflates it instead); confusing dip direction (270° here) with dip azimuth of strike when writing the strike/dip notation.

Source: Goodman, engineering-geology mapping methods (strike/dip, three-point problem, apparent dip); Marshak, Earth: Portrait of a Planet, Ch. 23 (plate boundaries).

Question 3 of 4: 3-Point Problem and Tectonic Plates

Question 3: 3-Point Problem and Tectonic Plates (20 marks)

23. Strike, dip and outcrop trace of the coal seam

24. Tectonic plates and boundaries (Figure Q24)