Chapter 12.3 — Exercise 12.2 — Two Triangle Problems
Problems based on figures with two triangles. This is Lesson 3 of 3 in Chapter 12: Applications of Trigonometry.
When One Triangle Isn't Enough Information
Every problem in Exercise 12.2 needs two separate right triangles, solved together, because a single angle and a single triangle no longer pin down a unique answer. The two triangles almost always share exactly one unknown, which is what makes solving them jointly possible — one triangle's equation gets substituted straight into the other's, exactly the way a small system of two equations in two unknowns would normally be solved.
A Tower, Viewed From Two Points
A tower's angle of elevation is 60° from one point, and 30° from a second point 10 m further back, in line with the tower's base. Both the tower's height h and the first point's distance x from the tower are unknown.
tan60° = h/x ⟹ h = x√3 ...(1)
tan30° = h/(x+10) ⟹ 1/√3 = x√3/(x+10) ⟹ x+10 = 3x ⟹ x = 5
h = 5√3 m ≈ 8.66 m, width of the road = x = 5 mNeither triangle alone can solve this problem — the closer triangle has two unknowns (h and x) with only one equation, and so does the farther one. It's only once both equations are combined, substituting one into the other, that a single unknown (x) becomes solvable, after which h follows immediately.
Notice, too, that both triangles here share not just one unknown but two — h and x both appear in each equation, just combined differently. That's a genuinely richer sharing than most of the problems later in this exercise, where usually only a single quantity (a height, or a base distance) carries over between the two triangles.
Walking Toward a Temple
A 1.5 m tall boy watches a 30 m temple; as he walks closer, the elevation angle rises from 30° to 60°. The height above his own eye level is 30−1.5=28.5 m throughout — only the horizontal distance changes as he walks.
tan60° = 28.5/x ⟹ x = 28.5/√3 = 9.5√3 m (closer distance)
tan30° = 28.5/(x+y) ⟹ x+y = 28.5√3 m (farther distance)
distance walked, y = 28.5√3 − 9.5√3 = 19√3 ≈ 32.91 mThe boy's height plays exactly the same role here as the observer's height did in Exercise 12.1's palm tree problem — subtracted once at the very start, then never touched again by either triangle. Both triangles work entirely with the reduced 28.5 m height above eye level; only the very last step, if the problem asked for a height rather than a distance, would need 1.5 m added back.
It's worth noticing which quantity actually gets asked for here: not either triangle's individual distance, but the difference between them — how far the boy actually walked between his two observation points. That difference, y=x+y minus x, only comes from computing both triangles' full distances first and subtracting; there's no shortcut that reaches the walked distance without finding both individual distances along the way.
Six More Two-Triangle Problems, Compressed
| Setup | Key relationship | Result |
|---|---|---|
| Statue on a 2 m pedestal; elevation 60° to statue top, 45° to pedestal top | tan45° finds the 2 m distance; tan60° then finds statue+pedestal together | statue height = 2(√3−1) ≈ 1.46 m |
| Building to cell tower: 60° elevation to tower top, 45° depression to tower foot, 7 m apart | 45° depression fixes the building's own height at 7 m (equal legs); 60° then finds the rest | tower height = 7(√3+1) ≈ 19.12 m |
| 18 m wire re-tied from 30° to 60° elevation, same pole | sin30° fixes the pole height at 9 m; sin60° finds the new, shorter wire | 7.608 m of wire cut off |
| Tower (30 m) and building see each other's tops at 60° and 30° | tan60° from the tower's triangle finds the shared base distance; tan30° then finds the building | building height = 10 m |
| Two equal-height poles, 120 ft apart, seen at 60° and 30° from a point between them | Both triangles' heights set equal, since the poles match | height = 30√3 ft; point sits 30 ft and 90 ft from each pole |
| Tower/building ratio: 30° and 60° elevations to each other's tops | Both triangles share the same base distance BD, eliminated by substitution | ratio 1 : 3 |
Every row shares the identical underlying shape: two triangles, one shared unknown (a distance or a height), and one substitution step that collapses two equations into one solvable line. The building-and-tower ratio problem is worth a second glance for what it never actually finds — neither individual height is ever computed, only the ratio between them, since the shared base distance cancels out of the final relationship entirely before either height needs a specific numeric value.
A Height From Two Complementary Angles
A tower's elevation angles measured from two separate points, 4 m and 9 m away, are complementary — meaning together they add to exactly 90°. Letting the nearer angle be θ makes the farther one 90°−θ automatically.
tanθ = height/4 ...(1)
tan(90°−θ) = cotθ = height/9 ...(2)
Multiplying (1)×(2): tanθ·cotθ = height²/36 ⟹ 1 = height²/36 ⟹ height = 6 mMultiplying the two equations together, rather than solving either one on its own, is the genuinely elegant move here — tanθ·cotθ always equals exactly 1, regardless of what θ actually is, so the entire angle disappears from the calculation without ever needing to be found. The tower's height turns out to be exactly the geometric mean of the two given distances, √(4×9)=6 — a clean result specific to this complementary-angle setup, not something that would happen for two arbitrary, non-complementary elevation angles.
A Jet Plane's Speed
A jet flies at a constant height of 1500√3 m. Its elevation angle from a fixed ground point drops from 60° to 30° over 15 seconds.
tan60° = 1500√3/(distance at 60°) ⟹ distance = 1500 m
tan30° = 1500√3/(distance at 30°) ⟹ distance = 4500 m
Distance flown = 4500 − 1500 = 3000 m in 15 s ⟹ speed = 200 m/s = 720 km/hrBoth triangles share the identical height throughout — the plane never changes altitude — which is exactly why this problem reduces to finding two horizontal distances and subtracting, rather than anything more complicated. Converting the final speed from metres per second to kilometres per hour (×18/5) is worth doing as a matter of habit whenever a problem's given units and expected-answer units don't already match.
This problem is also worth noticing as the only one in the entire chapter that finishes with a rate rather than a length, an area, or a ratio — distance divided by time, computed only after every single trigonometric step has already finished. The two triangles do all of the geometric work; the final speed calculation itself is ordinary arithmetic with nothing trigonometric left in it at all.
The Chapter, Complete
Applications of Trigonometry took every ratio and identity built up across Chapter 11 and finally pointed them at genuine physical situations: the introduction defined elevation, depression, and the three conventions every problem since has relied on, Exercise 12.1 solved scenarios needing one triangle each, and this exercise closed with scenarios needing two triangles solved jointly. The next chapter, Probability, leaves triangles behind entirely, turning instead to how likely an uncertain event actually is.