Multi-Step Trigonometry Problems
Combine Pythagoras with trig in problems that need two or more steps: surveyor-style situations where the missing measurement requires intermediate calculations.
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You have a right triangle, but only ONE side and the right angle are given, no angle measurements. Can you find the other two sides? Why not (or how, given extra info)?
Real-world problems rarely give you exactly what you need. Often you must do one trig calculation to get an intermediate length, then a SECOND calculation to find the final answer.
Typical patterns: (1) Use elevation/depression to find an intermediate distance, then use that to find a height. (2) Use Pythagoras to find a missing side, then use trig on that side. (3) Combine two triangles, solve the first, use its answer to solve the second.
Text description
- A horizontal ground line with a vertical side labelled height rising from it. A right-angle mark sits at the foot of the height.
- From the top of the height, two straight lines slope down to the ground on the right.
- The nearer line makes an angle of 35° with the vertical; its arc runs from the vertical to that line.
- The farther line makes an angle of 50° with the vertical; its larger arc runs from the vertical to that line.
- Below the ground, a dimension line labelled leg 1 runs from the foot of the height to where the 35° line lands, and a longer one labelled leg 2 runs from the foot of the height to where the 50° line lands.
Know
- Some problems require two or more trig steps
- Pythagoras can fill in a missing side that trig then operates on
- Intermediate answers should keep full precision
Understand
- How to break a complex problem into smaller sub-problems
- Why rounding mid-way causes inaccuracies
- How surveyors use base-distance measurements with two angles to find inaccessible heights
Can Do
- Solve two-step problems combining Pythagoras and trig
- Handle problems with two triangles sharing a common side
- Communicate the solution clearly with intermediate steps
Wrong: Trying to find the final answer in one step when you have only partial information.
Right: Break the problem into manageable steps. Find an intermediate length first.
Wrong: Rounding too early, carrying rounded numbers into the next step amplifies errors.
Right: Keep full calculator precision; round only the FINAL answer.
To find the height of a mountain that's not directly reachable, measure the elevation from TWO points along a known baseline.
Let $h$ = mountain height, $d$ = horizontal distance from the closer observation point. From point B: $h = d\tan\theta_2$. From point A (further by baseline $L$): $h = (d + L)\tan\theta_1$. Equating gives $d(\tan\theta_2 - \tan\theta_1) = L\tan\theta_1$, so $d = \dfrac{L\tan\theta_1}{\tan\theta_2 - \tan\theta_1}$, then $h = d\tan\theta_2$.
Sometimes you must use Pythagoras to find a side, then use that side as the input to a trig calculation.
| Situation | Order |
|---|---|
| Find diagonal of rectangle, then angle | Pythagoras (sides) → trig (angle) |
| Find leg of triangle from hyp + leg, then angle | Pythagoras → inverse trig |
| 3D problem (box, room) | Pythagoras (base diagonal) + Pythagoras (with height) → trig |
Watch Me Solve It · 3 examples
- 1Set up two equations$h = (d+100)\tan 25°$ and $h = d\tan 40°$ where $d$ = distance from B
- 2Equate and solve for $d$$d\tan 40° = (d+100)\tan 25°$, so $d(\tan 40° - \tan 25°) = 100\tan 25°$. $d = 46.6308/0.3727920 \approx 125.09$ m
- 3Compute height$h \approx 125.0852 \tan 40° \approx 104.96$ m
- 1Floor diagonal$d = \sqrt{8^2+6^2} = 10$ m (Pythagoras)
- 2Wire lengthWire = $\sqrt{10^2+3^2} = \sqrt{109} \approx 10.44$ m
- 3Angle with floor$\tan\theta = 3/10 = 0.3$, $\theta = \tan^{-1}(0.3) \approx 16.70°$
- 1Cliff height$h_c = 300\tan 22° \approx 121.21$ m
- 2Distance from cliff to seagullThe given horizontal distance and depression angle give a drop of $50\tan 18° \approx 16.25$ m below the cliff-top.
- 3Seagull height$121.21 - 16.25 = 104.96$ m above sea level.
Common Pitfalls
Strategy
- Sketch all triangles
- Identify intermediates
- Solve step-by-step
Tools
- Pythagoras for sides
- Trig ratios for angles
- Inverse trig for unknown angles
Surveyor method
- Two observation points
- Baseline known
- Two angles measured
Precision
- Keep decimals through
- Round only at end
- Verify with another method
How are you completing this lesson?
Brain Trainer · 4 problems
Four quick drills to lock in today's skill. Try each, then reveal the answer.
-
1 Rectangle 6 by 8. Diagonal?
$\sqrt{36+64}=10$.10 -
2 From 50 m: angle 20°. Height?
$50\tan 20° \approx 18.20$.$\approx 18.20$ -
3 Points A, B and the mountain foot lie on one straight line on level ground, with B between A and the foot. A and B are 50 m apart; the elevation is 30° from A and 45° from B. Find the mountain height.
Set up two eqs: $h = d\tan 45$ and $h = (d+50)\tan 30$. Solve $d \approx 68.30$, $h \approx 68.30$.$\approx 68.30$ m -
4 Box: 3$\times$4$\times$5. Space diagonal?
$\sqrt{9+16+25}=\sqrt{50}\approx 7.07$.$\approx 7.07$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. A rectangular box has dimensions 5 cm by 12 cm by 9 cm. (a) Find the diagonal of the base. (b) Find the space diagonal (corner to opposite corner). (c) Find the angle the space diagonal makes with the base.
Q7. From point A on flat ground, a tower top is at elevation 32°. From point B (60 m closer to the tower), the elevation is 50°. Find the height of the tower (2 d.p.).
Q8. A wire is anchored at the top of a flagpole and stretched tight to the ground 6 m from the base. The angle the wire makes with the ground is 65°. A second wire goes from the flagpole top to a point on the ground 10 m from the base. (a) Find the height of the flagpole. (b) Find the length of the first wire. (c) Find the angle the second wire makes with the ground.
Quick Check
1. B Keep precision.
2. B The points are 20 m apart. With $d$ the distance from B, $h=d\tan45^\circ=(d+20)\tan30^\circ$, so $h\approx27.32$ m.
3. C$\sqrt{225}=15$ (3-4-5 scaled).
4. A$\sqrt{169}=13$.
5. C$\cos^{-1}(0.3) \approx 72.5°$.
Show Your Working Model Answers
Q6 (3 marks): (a) Base diagonal $= \sqrt{25+144}=\sqrt{169}=13$ cm [1]. (b) Space diagonal $= \sqrt{25+144+81}=\sqrt{250}\approx 15.81$ cm [1]. (c) $\tan\theta = 9/13$, $\theta = \tan^{-1}(9/13) \approx 34.7°$ [1].
Q7 (3 marks): Let $d$ = distance from B. $h = d\tan 50°$ and $h = (d+60)\tan 32°$ [1]. Solve: $d(\tan 50° - \tan 32°) = 60\tan 32°$, $d \approx 37.4922/0.5668842 \approx 66.14$ m [1]. $h \approx 66.1372\tan 50° \approx 78.82$ m [1].
Q8 (4 marks): (a) $h = 6\tan 65° \approx 12.86$ m [1]. (b) Wire 1 length $= \sqrt{12.86^2 + 6^2} = \sqrt{201.4} \approx 14.19$ m (or $6/\cos 65° \approx 14.19$ m) [1]. (c) Second wire angle: $\tan\theta = 12.86/10 = 1.286$, $\theta \approx 52.1°$ [1] (1 d.p.). Note flagpole height carries through [1].
Three-step surveyor
From point A, the tower base is on bearing 075° and the angle of elevation to the tower top is 18°. The tower is at horizontal distance 200 m on this bearing. Point B is 80 m north of A. Find the tower height and the horizontal distance from B to the tower, then calculate the elevation angle from B. Assume both points and the tower base are at the same level.
Reveal solution
Tower height $= 200\tan 18° \approx 64.98$ m. Tower position from A: E = $200\sin 75° \approx 193.19$, N = $200\cos 75° \approx 51.76$. From B (80 m north of A): N = $51.76 - 80 = -28.24$, E = 193.19. Horizontal distance from B $= \sqrt{28.24^2 + 193.19^2} \approx 195.24$ m. The elevation from B is $\tan^{-1}(64.98/195.24) \approx 18.41^\circ$.
Multi-step
One step at a time
Sketch first
Draw all triangles
Label intermediates
Give them letters
Keep precision
Round at the END
Surveyor method
Two angles + baseline
3D Pythagoras
$\sqrt{a^2+b^2+c^2}$
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