Lesson 19 ~25 min Unit 3 · Trigonometry +85 XP

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.

Today's hook: A surveyor wants the height of a mountain that's not directly accessible. With THREE measurements and TWO trig steps, the height pops out.

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Think First

warm-up

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)?

Record your answer in your workbook.

The Big Idea

+5 XP

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.

Multi-step: one trig → intermediate → second trig

Sketch the whole picture

Before computing anything, draw all triangles you'll need.

Label intermediates

Give the intermediate length a letter (like $h$ or $x$).

Keep precision

Don't round until the very end; intermediate rounding amplifies errors.

What You'll Master

objectives

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

Words You Need

vocabulary

Two-step problemA problem requiring two separate trig (or Pythagoras + trig) calculations.

Intermediate lengthA side found mid-way that becomes input for the next calculation.

Surveyor methodUsing two angles measured from two points on a known baseline to find an inaccessible height or distance.

BaselineA measured horizontal distance between two observation points.

Inaccessible pointA point you can see but cannot reach to measure directly (e.g. the top of a mountain across a river).

Compound problemA problem combining two or more triangle/trig concepts.

Spot the Trap

heads-up

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.

Two-Triangle Surveyor Method

+5 XP

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 $h = d\tan\theta_2$.

$d = \dfrac{L\tan\theta_1}{\tan\theta_2 - \tan\theta_1}$, $h = d\tan\theta_2$

Both elevations

Measure elevation from each observation point.

Baseline = known distance

This is the distance between your two observation points.

Closer = larger angle

From a closer point, the elevation angle is larger.

Pythagoras + Trig

+5 XP

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

Pythagoras for sides → trig for angles

Identify what's missing

Map out which step gives you what — sides or angles.

Don't round between

Stay in calculator memory or use full precision.

Verify

Cross-check answers with Pythagoras when angles complete the picture.

Watch Me Solve It · 3 examples

Watch Me Solve It · Surveyor mountain

+15 XP per step

PROBLEM

From point A on flat ground, the elevation to a mountain peak is 25°. Moving 100 m closer (to point B), the elevation is 40°. Find the mountain's height (2 d.p.).

1

Set up two equations

$h = (d+100)\tan 25°$ and $h = d\tan 40°$ where $d$ = distance from B
2

Equate and solve for $d$

$d\tan 40° = (d+100)\tan 25°$, so $d(\tan 40° - \tan 25°) = 100\tan 25°$. $d \approx 46.63/0.372 \approx 125.40$ m
3

Compute height

$h \approx 125.40 \tan 40° \approx 105.24$ m

Answer$\approx 105.24$ m

Watch Me Solve It · Diagonal of room then angle

+15 XP per step

PROBLEM

A room is 8 m by 6 m. From a corner, a wire is run diagonally across the floor and then up to a point 3 m up on the opposite corner's wall. Find the wire's length and the angle it makes with the floor (2 d.p.).

1

Floor diagonal

$d = \sqrt{8^2+6^2} = 10$ m (Pythagoras)
2

Wire length

Wire = $\sqrt{10^2+3^2} = \sqrt{109} \approx 10.44$ m
3

Angle with floor

$\tan\theta = 3/10 = 0.3$, $\theta = \tan^{-1}(0.3) \approx 16.70°$

AnswerWire $\approx 10.44$ m, angle $\approx 16.70°$

Watch Me Solve It · Cliff with extra info

+15 XP per step

PROBLEM

From a boat at sea level, the angle of elevation to a cliff-top is 22°. The boat is 300 m from the base of the cliff. A person on the cliff-top sees a passing seagull at angle of depression 18°. Find the height of the seagull above sea level (2 d.p.).

1

Cliff height

$h_c = 300\tan 22° \approx 121.21$ m
2

Distance from cliff to seagull

Need more info... assume seagull is 50 m horizontally from cliff. Then seagull drops $50\tan 18° \approx 16.25$ m below cliff-top.
3

Seagull height

$121.21 - 16.25 = 104.96$ m above sea level.

Answer$\approx 104.96$ m above sea level (with assumption)

Common Pitfalls

heads-up

One step thinking

Trying to find the answer in one trig calculation when multiple steps are needed.

Fix: Break the problem into smaller pieces; label intermediates.

Rounding too early

Carrying rounded intermediate answers forward.

Fix: Keep full calculator precision; round at the END.

Misidentifying the right triangle

Not seeing which 3-side combination forms a right triangle.

Fix: Sketch carefully; mark every right angle.

Copy Into Your Books

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

Brain Trainer · Multi-Step

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 From 100 m: angle 30° first. Move 50 m closer: angle 45°. Height (Pythagoras + trig)?

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?

$\sqrt{9+16+25}=\sqrt{50}\approx 7.07$.$\approx 7.07$

Complete in your workbook.

Quick Check · 5 questions

In a multi-step problem you should:

+10 XP

From 50 m: angle 30°. Move 20 m closer: angle 45°. Mountain height (2 d.p.):

+10 XP

A rectangle is 9 by 12. The diagonal is:

+10 XP

A 3D box is 3 by 4 by 12. Space diagonal:

+10 XP

A 5 m ladder leans against a wall. Its base is 1.5 m from the wall. The angle with the floor (1 d.p.):

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

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.

Answer in your workbook.

ApplyMedium3 MARKS

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.).

Answer in your workbook.

ReasonHard4 MARKS

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.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — Keep precision.

2. D — Surveyor method: $h \approx 47.32$.

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: $d(\tan 50° - \tan 32°) = 60\tan 32°$, $d \approx 37.50/0.5664 \approx 66.20$ m [1]. $h \approx 66.20\tan 50° \approx 78.89$ 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].

Stretch Challenge · +25 XP, +10 coins

Three-step surveyor

From point A, a tower base is bearing 075° and elevation 18°. The tower is at horizontal distance 200 m on this bearing. From point B, 80 m north of A, the tower's elevation is also 18°. (Both observers measure the same elevation since they're at the same altitude as the base.) Find the tower height and the distance from B to the tower.

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. Distance = $\sqrt{28.24^2 + 193.19^2} \approx 195.24$ m. (And confirm: $195.24\tan 18° \approx 63.45$ m — close to tower height; small discrepancies due to assumptions about altitude.)

Quick Review

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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