Lesson 8 ~25 min Unit 3 · Trigonometry +85 XP

Cosine Ratio — Finding a Side

Use $\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$ to find the adjacent (adj = hyp $\times$ cos) or the hypotenuse (hyp = adj $\div$ cos) when an angle and one side are known.

Today's hook: A 10 m wire supports a tower. The wire makes a 65° angle with the ground. The distance from the tower base to the anchor is the adjacent side — how far is the anchor from the base?

0/5QUESTS

Printable Worksheets

Print or save as PDF — or build a custom worksheet from any module's questions.

Build Apply Master Build custom

Think First

warm-up

You know hypotenuse = 10 m and the angle = 65°. You want the adjacent. Which trig ratio links adj and hyp? Hint: CAH.

Record your answer in your workbook.

The Big Idea

+5 XP

Cosine connects the adjacent side and the hypotenuse. Whenever the diagram involves the leg NEXT to $\theta$ (not across from it) and the hypotenuse, use cos.

$\cos\theta = $ adj/hyp. Rearranged: adj = hyp $\times \cos\theta$ (when hyp is known) and hyp = adj $\div \cos\theta$ (when adj is known). Round to 2 d.p.

adj = hyp $\times \cos\theta$ or hyp = adj $\div \cos\theta$

CAH

Cosine = Adjacent / Hypotenuse.

Adj is next to $\theta$

The leg $\theta$ touches, but not the hypotenuse.

Pair with sin

sin and cos differ only in whether you use opp or adj.

What You'll Master

objectives

Know

$\cos\theta = $ adj/hyp
adj $=$ hyp$\cdot \cos\theta$
hyp $=$ adj$/\cos\theta$

Understand

When to choose cos (when adj and hyp are involved)
Why cos of an acute angle is between 0 and 1
Why cos of 90° equals 0 and cos of 0° equals 1

Can Do

Decide between sin, cos and tan from a diagram
Solve for adj or hyp using cosine
Verify answers using Pythagoras as a cross-check

Words You Need

vocabulary

$\cos\theta$Cosine of $\theta$: ratio adj/hyp.

AdjacentThe leg next to $\theta$ but not the hypotenuse.

AnchorThe point on the ground where a supporting cable is fixed.

Guy wireA wire that supports a tall structure (tower, mast) at an angle.

Horizontal distanceThe flat (ground-level) distance — usually the adjacent in vertical-tower problems.

$\cos 0° = 1$At 0°, the adjacent equals the hypotenuse, so the ratio is 1.

Spot the Trap

heads-up

Wrong: Using sin instead of cos when the adjacent is involved — gives a completely different answer.

Right: Check which leg is involved. Adj + hyp → cos. Opp + hyp → sin.

Wrong: “adj = hyp$/\cos\theta$.” Wrong — that gives hyp from adj, not adj from hyp.

Right: To find adj from hyp: multiply by cos. To find hyp from adj: divide by cos.

Choosing Between Sin and Cos

+5 XP

Sin and cos look similar but use different legs. The decision rule is simple.

Sides involved	Use	Formula
opp + hyp	$\sin$	$\sin\theta = $ opp/hyp
adj + hyp	$\cos$	$\cos\theta = $ adj/hyp
opp + adj	$\tan$	$\tan\theta = $ opp/adj

Pick the ratio matching the two sides involved

List sides first

Write which two sides are known/wanted, then pick.

Hyp + adj → cos

If you see hyp and adj only, the ratio is cosine.

Same method

Same set up → rearrange → compute as sin lessons.

Cosine in Practical Problems

+5 XP

Cosine often appears in problems about supporting cables, ladder bases, sliding doors, ramps measured along the ground — anything where the HORIZONTAL distance matters.

A guy wire of length $L$ at angle $\theta$ to the ground: ground distance from anchor to base = $L\cos\theta$. A ladder of length $L$ leaning at angle $\theta$: foot-of-ladder distance from wall = $L\cos\theta$. A car driving up a $\theta$-degree slope for distance $L$: horizontal distance covered = $L\cos\theta$.

Horizontal distance = (slant) $\times \cos\theta$

Horizontal = adj

In tower/ladder problems, the horizontal (ground) distance is the adjacent.

Vertical = opp

The vertical height is the opposite (use sin for that).

Sketch helps

A 3-second sketch saves you from picking the wrong ratio.

Watch Me Solve It · 3 examples

Watch Me Solve It · Guy wire base distance

+15 XP per step

PROBLEM

A 10 m wire supports a tower. The wire makes a 65° angle with the ground. How far from the tower's base is the anchor point (2 d.p.)?

1

Set up

hyp = 10, adj = ?, $\theta = 65°$
2

Apply CAH

$\cos 65° = $ adj/10, so adj $= 10\cos 65°$
3

Compute

adj $\approx 10 \times 0.4226 \approx 4.23$ m

Answer$\approx 4.23$ m

Watch Me Solve It · Find the hypotenuse from adj

+15 XP per step

PROBLEM

In a right triangle, adj = 15 and $\theta = 40°$. Find the hypotenuse (2 d.p.).

1

Set up

$\cos 40° = 15/$hyp
2

Rearrange

hyp $= 15/\cos 40°$
3

Compute

hyp $\approx 15/0.766 \approx 19.58$

Answerhyp $\approx 19.58$

Watch Me Solve It · Ladder foot

+15 XP per step

PROBLEM

A 6 m ladder leans against a wall at 70° to the ground. How far is the foot of the ladder from the wall (2 d.p.)?

1

Identify

hyp = 6 (ladder), adj = distance to wall, $\theta = 70°$
2

Apply

adj $= 6\cos 70°$
3

Compute

$\approx 6 \times 0.342 \approx 2.05$ m

Answer$\approx 2.05$ m from the wall

Common Pitfalls

heads-up

Confusing sin and cos

Using sin when the adjacent is involved — wrong leg.

Fix: List the two sides involved first: opp+hyp $\to$ sin, adj+hyp $\to$ cos.

Wrong rearrangement

Writing adj = hyp / cos instead of adj = hyp $\times$ cos.

Fix: From cos = adj/hyp, multiply both sides by hyp.

Wrong angle reference

Mixing up which angle is $\theta$.

Fix: Mark $\theta$ on your diagram. The adj is the leg touching $\theta$.

Copy Into Your Books

CAH

$\cos\theta = $ adj/hyp
adj $=$ hyp$\cdot \cos\theta$
hyp $=$ adj$/\cos\theta$

When to use

Sides involved: adj + hyp
Includes angle $\theta$
Horizontal-distance problems

Common contexts

Guy wires
Ladders
Slope horizontal-run

Range

$0 \le \cos\theta \le 1$ for acute $\theta$
$\cos 0° = 1$
$\cos 90° = 0$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Cosine Side Hunt

4 problems

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

1 hyp = 10, $\theta = 60°$. Find adj.

adj $= 10\cos 60° = 10 \times 0.5 = 5$.$= 5$
2 adj = 6, $\theta = 45°$. Find hyp (2 d.p.).

hyp $= 6/\cos 45° \approx 6/0.7071 \approx 8.49$.$\approx 8.49$
3 hyp = 14, $\theta = 30°$. Find adj (2 d.p.).

adj $= 14\cos 30° \approx 14 \times 0.866 \approx 12.12$.$\approx 12.12$
4 adj = 9, $\theta = 35°$. Find hyp (2 d.p.).

hyp $= 9/\cos 35° \approx 9/0.819 \approx 10.99$.$\approx 10.99$

Complete in your workbook.

Quick Check · 5 questions

$\cos\theta$ equals:

+10 XP

hyp = 8, $\theta = 60°$. The adjacent is:

+10 XP

adj = 12, $\theta = 30°$. The hypotenuse is (2 d.p.):

+10 XP

You know the hyp and the angle, want the side NEXT to the angle. Use:

+10 XP

A 25 m guy wire from a tower top hits the ground at 70°. Ground distance from tower base (2 d.p.):

+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) hyp = 20, $\theta = 40°$, find adj. (b) adj = 6, $\theta = 55°$, find hyp. (c) hyp = 8.5, $\theta = 28°$, find adj.

Answer in your workbook.

ApplyEasy2 MARKS

Q7. A 12 m guy wire from the top of a flagpole makes a 60° angle with the ground. How far from the base is the wire anchored?

Answer in your workbook.

ReasonHard4 MARKS

Q8. A 4 m ladder leans against a wall. Safety code requires the foot of the ladder to be no closer than 1 m from the wall, and no further than 1.5 m. Find the range of safe angles (to the nearest degree) the ladder can make with the ground.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — CAH.

2. A — $8\cos 60° = 4$.

3. C — $12/\cos 30° \approx 13.86$.

4. B — adj + hyp → cos.

5. D — $25\cos 70° \approx 8.55$.

Show Your Working Model Answers

Q6 (3 marks): (a) adj $= 20\cos 40° \approx 15.32$ [1]. (b) hyp $= 6/\cos 55° \approx 10.46$ [1]. (c) adj $= 8.5\cos 28° \approx 7.50$ [1].

Q7 (2 marks): adj $= 12\cos 60°$ [1] $= 12 \times 0.5 = 6$ m [1].

Q8 (4 marks): $\cos\theta = $ adj/4 [1]. For adj $= 1$: $\cos\theta = 0.25$, $\theta \approx 76°$ [1]. For adj $= 1.5$: $\cos\theta = 0.375$, $\theta \approx 68°$ [1]. Safe range: 68° to 76° from the ground [1].

Stretch Challenge · +25 XP, +10 coins

Sliding ladder — cos vs sin

A 5 m ladder leans at 70° against a wall. It is pulled out so it now leans at 60°. By how much (2 d.p.) does the foot of the ladder move away from the wall?

Reveal solution

Initial adj $= 5\cos 70° \approx 1.71$. New adj $= 5\cos 60° = 2.5$. The foot moves $2.5 - 1.71 = 0.79$ m further from the wall.

Quick Review

Cosine ratio

$\cos\theta = $ adj/hyp

Find adj

adj $=$ hyp$\cdot \cos\theta$

Find hyp

hyp $=$ adj$/\cos\theta$

When to use

adj + hyp involved

Common context

Horizontal distance from slope

Range

$0 \le \cos\theta \le 1$ for acute $\theta$

Your Badges

0 of 6

First Steps

3-Day Streak

3 in a Row

Lesson Ace

Stretch Seeker

Daily Warrior

Mark lesson as complete

Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.

Unit overview · Maths subject page