Two angles in one figure, and the pair of equations they hand you

What to assume they know

• Picking the one ratio that links what you know to what you want — selecting a ratio from the known and wanted sides
• Looking up versus looking down: elevation and depression — moving a depression angle to the far vertex before writing a ratio
• Solving two simultaneous equations by substitution, and eliminating a shared unknown
• Rationalising a denominator of the form √3 − 1 by multiplying above and below by √3 + 1
• That opposite sides of a rectangle are equal
• Exact values of the ratios at 30°, 45° and 60°, and that tan 30° and tan 60° are reciprocals

What they should be able to do

• Recognise from a problem statement that no length inside either triangle has been given, and that a second angle is therefore doing structural work
• Draw a two-sighting figure so that the two right triangles share their vertical side, or share their base, as the situation requires
• Write one ratio equation per triangle, in the same two unknowns
• Eliminate the unmeasurable length between the two equations and solve for the wanted one
• Say, for a given problem, whether the number supplied is a plain length or a difference of two lengths, and why several of the hardest items in this chapter supply only a difference
• Rationalise a surd denominator and report the height as an exact expression
• Check a two-angle figure for consistency by confirming that the target further from the observer's own horizontal carries the larger angle, and say why that reads as "higher" for elevation and "lower" for depression
• Identify the case in which the unknown height cancels entirely, and say what that means physically

Where it usually goes wrong

• "Two angles means two separate problems." They are one problem with two equations. Solving each triangle alone gets nowhere, because neither has a known side.
• "The 40 m in Example 5 is the shadow." It is how much longer one shadow is than the other. Reading it as a shadow length gives a different and wrong equation, and it is the single commonest error on this example.
• "You need to know the base distance." You never need it given. In Q10, Q11 and Q15 it is neither given nor asked for. Example 5 is the instructive variant: the base is not given either, but the working does end up computing it, at 20 m, purely as a step towards the height. Either way it is machinery, not the answer.
• "The two angles belong to the same triangle." They belong to two triangles that overlap. Draw them apart once before drawing them together.
• "A bigger angle means a further station." The opposite. Walking towards something raises the angle to its top. Q6 and Q14 both turn on this and students reliably invert it.
• "Rationalising is a presentation rule." In Example 6 it is what turns an opaque quotient into a height you can read as a sum of two terms, one of which is the short building's own 8 m.
• "In the balloon problem the 88.2 m is the height above her eyes." It is above the ground. Her 1.2 m comes off first, exactly as in the ground-level correction from the first module.
• "The car problem needs the tower's height." It cancels. If a student cannot see why, they have not understood what elimination is for.

Questions to check understanding

• A shadow lengthening or shortening between two stated Sun altitudes — find the height
• Two stations a stated distance apart on one line with the object — find the height and the near distance
• A stacked pair of targets from one station — find the upper section's length
• Two depressions from one elevated point — find a height, a separation, or both
• An observer of stated height walking towards a building between two elevations — find the distance walked
• A timed pair of sightings at uniform speed — find a speed or a remaining time, and state why the height is not needed
• Show that a two-angle configuration is inconsistent if the larger angle is attached to the further station

Examples worth working on the board

Values marked verified are an added algebra on data printed inside pp. 133–143.

• Example 5, the purest case (pp. 138–139, Fig. 9.8). Inputs: a tower on level ground; its shadow is 40 m longer when the Sun stands at 30° than when it stands at 60°; the tower's height is wanted. No length in either triangle is given — the 40 m is the gap between two shadow tips. Fig. 9.8 letters the tower's top A and its foot B, the near shadow tip C and the far tip D, prints 30° at D and 60° at C, and marks 40 m along DC with a double arrow. Verified, an added algebra: with AB = h and BC = x, the 60° triangle gives √3 = h/x and the 30° triangle gives 1/√3 = h/(x + 40). Substituting h = x√3 into the second yields 3x = x + 40, so x = 20 and h = 20√3 ≈ 34.64 m. The moment to slow down on is which quantity leaves and which stays: it is the height that gets substituted out, and the near shadow x — never wanted, never measured — is what the resulting equation is solved for, at 20 m, after which the height is recovered from it. The book's printed working on p. 139 does the same and states x = 20 outright. Nothing is struck through; the unwanted length is found on the way to the wanted one. If the explanation wants an unknown to visibly disappear, the place for it is Exercise 9.1 question 15, where the height really does cancel.
• Example 4, two targets from one station (pp. 137–138, Fig. 9.7). Inputs: a building 10 m tall; from a ground point P its top is at 30° and the top of a flag hoisted above it is at 45°; the flag's length and the distance from P to the building are wanted, taking √3 = 1.732. Fig. 9.7 letters the building's foot A, its top B, the flag's top D, the station P, and prints both angles at P; the flag is drawn as a small pennant at D. Verified: from tan 30° = 10/AP, AP = 10√3 ≈ 17.32 m. Then with DB = x, tan 45° = (10 + x)/(10√3) gives 10 + x = 10√3, so x = 10(√3 − 1) ≈ 7.32 m. Note the shape of the answer: the flag's length is the difference of two heights, just as Example 5's given was the difference of two distances.
• The consistency check (mine, usable on Figs. 9.7, 9.8, 9.9, 9.12, 9.13). State it in the only form that survives both kinds of angle: from a fixed station at a fixed horizontal distance, the target lying further from the observer's own horizontal carries the larger angle. Looking up, further from the horizontal means higher, so a higher target reads larger; looking down, it means lower, so the rule inverts. Fig. 9.7 is an elevation pair and obeys the upward form — the flag top D sits above the building top B and carries 45° against 30°. Fig. 9.8 fixes the target instead and varies the station: C is nearer the tower than D and carries 60° against 30°. Fig. 9.9 is the one that will catch a careless statement of this: it is a depression pair, the two targets sit at equal horizontal distance because the figure's rectangle makes them so, and the upper one carries 30° while the lower carries 45°. A student taught "higher means larger" as a flat reflex will read Example 6 as mislabelled and swap the two angles, which changes the answer. Taught in the distance-from-the-horizontal form, the reflex still catches transposed labels and does not misfire here.
• Example 6, the same idea from above (pp. 139–140, Fig. 9.9). Inputs: an 8 m building is sighted twice from a taller building's roof — its own roof reads 30° below level, its base reads 45°. Lettering: tall building PC with P on top, short building AB with B on top, D on PC level with B, Q on the level ray at P. Verified: transferring the angles gives ∠PBD = 30° and ∠PAC = 45°; then BD = PD√3 and PC = AC; with AC = BD and DC = AB = 8, PD + 8 = PD√3, so PD = 8/(√3 − 1) = 4(√3 + 1) after rationalising, and the height is PC = 4(3 + √3) ≈ 18.93 m, equal to the separation AC. Be careful with the word share here, which Example 5's bullet uses for a literally common segment. The two triangles in this figure hold no horizontal segment in common at all — BD runs at the height of the short building's roof and AC runs along the ground, two different segments made equal by the rectangle. What they do literally share is a stretch of the vertical: PD is the upper part of PC. The contrast worth drawing is that Example 5 links its triangles through a common height while Example 6 links its through two equal horizontal distances, and the 8 m enters as a piece of the tall building's plumb line.
• Rationalising, worked in the open (p. 140). The book multiplies above and below by √3 + 1 and the denominator collapses to 2. Show the collapse; students treat rationalising as ritual until they watch the difference of squares do it.
• Example 7, sharing a perpendicular (pp. 140–141, Fig. 9.10). Inputs: a bridge deck sits 3 m above the banks; from a point on it the two banks show depressions of 30° and 45°; the river's width is wanted. Lettering: banks A and B, bridge point P with both angles, D the foot of the perpendicular, 3 m marked on the drawing. Verified: AD = 3/tan 30° = 3√3, BD = 3, width AB = 3(1 + √3) ≈ 8.20 m. Here the shared quantity is the 3 m and it is known, so nothing needs eliminating.
• Exercise 9.1's two-angle family (pp. 141–143), inputs only. Q6: a boy 1.5 m tall walks towards a 30 m building; the elevation to its top goes from 30° to 60°; the walked distance is wanted. Q10: a road 80 m wide with equal poles facing each other; from a point between them the two tops are at 60° and 30°; the poles' height and the two distances are wanted. Q11 (Fig. 9.12): a TV tower on a canal bank; from the point directly opposite it the top is at 60°, and from a point 20 m further back along the same line, 30°; the tower's height and the canal's width are wanted. Fig. 9.12 letters the tower's top A, its foot B, the near station C with 60°, the far station D with 30°, and marks 20 m along DC. Q13: from a 75 m lighthouse, two ships in line on the same side at depressions of 30° and 45°. Q14 (Fig. 9.13): a girl 1.2 m tall watches a balloon drifting level at 88.2 m above the ground; the elevation from her eyes falls from 60° to 30°; the balloon's travel is wanted. Fig. 9.13 draws the girl at the left with both angles at her eye, two balloons at the same height, and marks 88.2 m on the right-hand vertical down to ground level. Verified, working added here: Q6 → the rise above eye level is 28.5 m, the two distances are 28.5√3 and 9.5√3, so the walk is 19√3 ≈ 32.91 m. Q10 → 3x = 80 − x gives x = 20, so the poles are 20√3 ≈ 34.64 m and the point is 20 m and 60 m from them. Q11 → 3w = w + 20 gives a canal 10 m wide and a tower 10√3 ≈ 17.32 m. Q13 → 75(√3 − 1) ≈ 54.9 m. Q14 → the rise above her eyes is 87 m, the distances are 29√3 and 87√3, so the balloon travelled 58√3 ≈ 100.46 m. Every one of these is the same elimination.
• Q15, the case where the height cancels (pp. 142–143). Inputs: a car approaches a tower along a straight road at steady speed; from the top it is seen at 30°, and six seconds later at 60°; the remaining time to the foot is wanted. Verified: with height h, the two distances are h√3 and h/√3, so the six seconds cover h√3 − h/√3 = 2h/√3 while the remainder is h/√3 — exactly half. The answer is 3 s and h never has to be known. This is the strongest possible statement of the topic's thesis and belongs at the end of the explanation.

Figures to have open

• Fig. 9.8 (p. 138): tower AB, shadow tips C and D, 30° at D, 60° at C, 40 m marked along DC. The most important single figure in this topic.
• Fig. 9.7 (printed on p. 138; p. 137 carries only the text reference and the statement of Example 4): station P with both angles, building AB, flag top D. Standard schematic; the pennant is decoration.
• Fig. 9.9 (p. 139): P, Q, B, D, A, C with the rectangle ABDC visible.
• Fig. 9.10 (p. 140): the bridge, P, the 3 m perpendicular to D, banks A and B.
• Fig. 9.12 (p. 142): tower AB on the far bank, stations C and D, 20 m marked between them, 60° at C and 30° at D.
• Fig. 9.13 (p. 142): the girl, two balloon positions at one level, both angles at her eye, 88.2 m marked from ground.
• A purpose-built movement for section 4: the two equations side by side, with the shared unknown highlighted in both and then removed. This is an added figure and it is what the topic is for.

Where this sits in the book

• NCERT Class X Mathematics, Chapter 9 "Some Applications of Trigonometry", §9.1 "Heights and Distances", Examples 4 to 7, pp. 137–141.
• Figs. 9.7 (p. 138), 9.8 (p. 138), 9.9 (p. 139), 9.10 (p. 140), 9.12 and 9.13 (p. 142).
• Exercise 9.1 questions 6 to 15, pp. 141–143.
• §9.2 Summary, p. 143, point 2 — a single line covering both heights and the distance between two remote objects.