Lesson 9 ~25 min Unit 3 · Trigonometry +85 XP

Tangent Ratio — Finding a Side

Use $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$ to find the opposite (opp = adj $\times$ tan) or the adjacent (adj = opp $\div$ tan) when an angle and a leg are known. No hypotenuse needed!

Today's hook: Standing 25 m from a building, you look up at the top at an angle of 40°. How tall is the building? You can solve this without knowing your line of sight (hypotenuse) — tan needs only the two legs.

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Think First

warm-up

You're 30 m from the base of a tree, looking up at the top at 28°. Which two sides of the right triangle do you know/want? Neither is the hypotenuse — which ratio fits?

Record your answer in your workbook.

The Big Idea

+5 XP

Tangent links the two legs of a right triangle — the opposite and the adjacent. It is the ratio of choice when the hypotenuse is not given AND not wanted.

$\tan\theta = $ opp/adj. Rearrange to opp $= $ adj $\cdot \tan\theta$ (when adj known) or adj $= $ opp $/ \tan\theta$ (when opp known). The hypotenuse appears nowhere — tan is the leg-only ratio.

opp = adj $\cdot \tan\theta$ or adj = opp $/ \tan\theta$

TOA

Tangent = Opp/Adj.

No hyp

Use tan when the problem ignores the hypotenuse.

Can exceed 1

Unlike sin and cos, tan can be much greater than 1 — e.g. $\tan 80° \approx 5.67$.

What You'll Master

objectives

Know

$\tan\theta = $ opp/adj
Rearranged: opp $=$ adj$\cdot \tan\theta$ and adj $=$ opp$/\tan\theta$
Tangent ratios can exceed 1

Understand

When to use tan instead of sin/cos
Why tan is undefined at 90°
Why looking up at a steep angle gives a large tan value

Can Do

Choose tan when no hypotenuse is involved
Solve for opp or adj using tan
Apply tan in height/distance problems

Words You Need

vocabulary

$\tan\theta$Tangent of $\theta$: ratio opp/adj.

Both legsThe two non-hypotenuse sides. Tan uses only these.

Steep angleAn angle close to 90°. The tangent becomes very large.

$\tan 45° = 1$When opp = adj, the triangle is isosceles and tan equals 1.

Tower / building problemA vertical structure of unknown height, viewed from a known horizontal distance at a known angle.

Looking upThe angle of elevation is between the horizontal (adjacent) and the line of sight (hypotenuse).

Spot the Trap

heads-up

Wrong: “tan needs the hypotenuse.” No — tan is the only ratio that ignores the hypotenuse.

Right: Tan uses opp and adj only. Great when no hyp is mentioned.

Wrong: Writing tan = adj/opp — reversed.

Right: TOA — opp is on TOP. $\tan\theta = $ opp/adj.

Choosing Tan

+5 XP

Tan is the right ratio whenever the diagram involves both legs and no hypotenuse. If you see a tower height and a ground distance, that's tan.

Decision check: Is the hypotenuse mentioned or needed? If NO → use tan. If yes → use sin or cos depending on which leg is involved.

No hyp → use tan

Two legs only

Tan handles opp and adj — no hyp.

Height + ground

Vertical height + horizontal distance → classic tan setup.

$\tan^{-1}$ later

Used in reverse to find angles (next lessons).

Tangent in Real Problems

+5 XP

Tan dominates tower-height, building-height and slope-grade problems. The horizontal distance and the vertical drop are both legs.

Scenario	adj	opp
Building height	Ground distance	Building height
Cliff height	Horizontal to cliff base	Cliff height
Road grade	Horizontal run	Vertical rise

opp = adj $\cdot \tan\theta$ for heights from ground distance

Sketch + label

Always sketch the scenario, mark opp and adj clearly.

Steep = big tan

45° gives tan = 1. Steeper angles give tan > 1.

Include observer eye height

For real building height, add the observer's eye height to the answer.

Watch Me Solve It · 3 examples

Watch Me Solve It · Building height

+15 XP per step

PROBLEM

Standing 25 m from a building, you look up at its top at 40°. How tall is the building (2 d.p., ignoring eye height)?

1

Set up

adj = 25, opp = ?, $\theta = 40°$
2

Apply TOA

$\tan 40° = $ opp/25
3

Compute

opp $= 25\tan 40° \approx 25 \times 0.839 \approx 20.98$ m

Answer$\approx 20.98$ m

Watch Me Solve It · How far from the tower?

+15 XP per step

PROBLEM

A tower is 60 m tall. From an observation point at the same level as its base, the top is seen at 35° above horizontal. How far away is the observation point (2 d.p.)?

1

Set up

opp = 60, adj = ?, $\theta = 35°$
2

Apply

$\tan 35° = 60/$adj
3

Rearrange + compute

adj $= 60/\tan 35° \approx 60/0.7002 \approx 85.69$ m

Answer$\approx 85.69$ m

Watch Me Solve It · Cliff at 22° from 200 m away

+15 XP per step

PROBLEM

A cliff is seen at 22° above horizontal from a point 200 m from the cliff base. Find the cliff height (2 d.p.).

1

Set up

adj = 200, $\theta = 22°$, opp = ?
2

Apply

opp $= 200\tan 22°$
3

Compute

$\approx 200 \times 0.4040 \approx 80.81$ m

Answer$\approx 80.81$ m

Common Pitfalls

heads-up

Using sin or cos instead of tan

When the problem doesn't mention the hypotenuse, sin/cos require an extra step.

Fix: If neither hyp is given nor wanted, tan is the simplest choice.

Reversed ratio

Writing tan = adj/opp.

Fix: TOA — opposite ON TOP.

Forgetting observer height

Reporting the calculated opp as the building height when the question implies adding observer eye height.

Fix: Check the problem: if it asks for full height from ground, add the eye height of the observer.

Copy Into Your Books

TOA

$\tan\theta = $ opp/adj
opp $=$ adj$\cdot \tan\theta$
adj $=$ opp$/\tan\theta$

When to use

No hypotenuse
Height + ground problems
Both legs involved

Range

$\tan 0° = 0$
$\tan 45° = 1$
$\tan 90°$ undefined

Real contexts

Building/tower heights
Road grades
Tree heights from a distance

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Tangent Side Hunt

4 problems

Four quick drills to lock in today's skill. Try each, then reveal the answer.

1 adj = 10, $\theta = 45°$. Find opp.

opp $= 10\tan 45° = 10$.opp $= 10$
2 opp = 20, $\theta = 30°$. Find adj (2 d.p.).

adj $= 20/\tan 30° \approx 20/0.577 \approx 34.64$.$\approx 34.64$
3 adj = 8, $\theta = 60°$. Find opp (2 d.p.).

opp $= 8\tan 60° \approx 8 \times 1.732 \approx 13.86$.$\approx 13.86$
4 opp = 7, $\theta = 25°$. Find adj (2 d.p.).

adj $= 7/\tan 25° \approx 7/0.466 \approx 15.01$.$\approx 15.01$

Complete in your workbook.

Quick Check · 5 questions

$\tan\theta$ equals:

+10 XP

Standing 50 m from a tower, you see the top at 35°. Tower height (2 d.p., ignoring eye height):

+10 XP

Choose tan when:

+10 XP

A 30 m flagpole is seen at 20° above horizontal. Distance from observer to flagpole base (2 d.p.):

+10 XP

$\tan 45° = $?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Find the requested side to 2 d.p. (a) adj = 18, $\theta = 32°$, find opp. (b) opp = 15, $\theta = 60°$, find adj. (c) adj = 7.5, $\theta = 55°$, find opp.

Answer in your workbook.

ApplyEasy2 MARKS

Q7. From 25 m away, you look up at the top of a building at an angle of 40°. How tall is the building (ignoring observer height)?

Answer in your workbook.

ReasonHard4 MARKS

Q8. From a point 60 m due south of a cell tower, an observer looks up at the top at 18°. Including the observer's eye height of 1.65 m, find the total height of the cell tower above ground level (2 d.p.). Explain why eye height must be added.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — TOA.

2. B — $50\tan 35° \approx 35.01$.

3. A — Tan for legs-only problems.

4. D — $30/\tan 20° \approx 82.42$.

5. A — $\tan 45° = 1$.

Show Your Working Model Answers

Q6 (3 marks): (a) opp $= 18\tan 32° \approx 11.25$ [1]. (b) adj $= 15/\tan 60° \approx 8.66$ [1]. (c) opp $= 7.5\tan 55° \approx 10.71$ [1].

Q7 (2 marks): opp $= 25\tan 40°$ [1] $\approx 20.98$ m [1].

Q8 (4 marks): Height above eye level $= 60\tan 18° \approx 19.50$ m [2]. Total height above ground $= 19.50 + 1.65 = 21.15$ m [1]. The triangle's adjacent is measured at the observer's eye level, not the ground, so the calculated ‘height’ is from eye level. To get the height above ground we must add eye height [1].

Stretch Challenge · +25 XP, +10 coins

Two angles, one tower

An observer at point A sees the top of a tower at 30° above horizontal. Moving 20 m closer to the tower, the angle increases to 45°. Find the height of the tower (2 d.p.).

Reveal solution

Let $d$ = new distance and $h$ = height. $h = d\tan 45° = d$. Also $h = (d+20)\tan 30°$, so $d = (d+20)\tan 30°$. $d - d\tan 30° = 20\tan 30°$. $d(1-\tan 30°) = 20\tan 30°$. $d = 20\tan 30°/(1-\tan 30°) \approx 11.547/0.4226 \approx 27.32$ m. Height $\approx 27.32$ m.

Quick Review

Tangent ratio

$\tan\theta = $ opp/adj

Find opp

opp $=$ adj$\cdot \tan\theta$

Find adj

adj $=$ opp$/\tan\theta$

No hyp needed

Tan uses two legs only

Range

$\tan 45° = 1$, can exceed 1

Real use

Tower/building heights, slopes

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