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Module 2 · L15 of 22 ~55 min MS-M2 · MEDIUM ⚡ +95 XP available

Right-Angled Trigonometry: Finding Unknown Angles

To find an unknown angle, work out the ratio from the two known sides, then apply the inverse trig function. The calculator gives decimal degrees — convert to degrees and minutes for exam answers.

Today's hook — If $\sin\theta = 0.5$, you know $\theta = 30°$. But what if $\sin\theta = 0.73$? What operation would you use on your calculator? And is there always only one possible angle?

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Worksheets

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Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

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Think First — gut answer before the lesson

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If $\sin\theta = 0.5$, you probably know that $\theta = 30°$. But what if $\sin\theta = 0.73$? What operation would you use on your calculator? And what does the answer mean — is it always the only possible angle?

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Inverse trig — reversing the process

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When you know two sides of a right-angled triangle and need the angle, you apply the inverse trigonometric function. This is the "undo" operation for $\sin$, $\cos$, or $\tan$.

Step 1: Label O, A, H and identify which two sides you know.
Step 2: Write the trig ratio (e.g. $\sin\theta = O/H$).
Step 3: Evaluate the ratio (divide the two numbers).
Step 4: Apply the inverse: $\theta = \sin^{-1}(\text{result})$.
Step 5: Convert to degrees and minutes if required.

$\theta = \sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ — choose based on which two sides you know

Inverse sine — $\theta = \sin^{-1}(O/H)$

Use when O and H are known. Calculator: [SHIFT][SIN] or [SIN⁻¹].

Inverse cosine — $\theta = \cos^{-1}(A/H)$

Use when A and H are known. Always check: is the side adjacent to the angle or opposite?

Inverse tangent — $\theta = \tan^{-1}(O/A)$

Use when O and A are known — no hypotenuse needed. Most common in practical problems.

What you'll master

Know

Key facts

$\theta = \sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ of the appropriate ratio
How to convert decimal degrees to degrees and minutes
The sum of angles in a triangle $= 180°$; use to find the other angle
In a right-angled triangle, both acute angles are complements ($90° - \theta$)

Understand

Concepts

Why inverse trig is the "undo" operation for trig functions
When to use degrees-and-minutes vs decimal degrees
How angles of elevation and depression create equal alternate angles

Can do

Skills

Find an unknown angle using inverse trig on a calculator
Express answers in both decimal degrees and degrees/minutes
Find both acute angles of a right-angled triangle
Solve practical angle problems in elevation, depression, and bearing

Key terms

Inverse trigonometric functionThe "reverse" of a trig function: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ take a ratio value and return the angle.

Degrees and minutesAngles expressed as whole degrees (°) plus minutes (') where 1° = 60'; convert decimal part by multiplying by 60.

Angle of elevationThe angle measured upward from the horizontal line of sight to an object above; always creates a right-angled triangle.

Angle of depressionThe angle measured downward from the horizontal line of sight to an object below; equal to the angle of elevation at the lower point (alternate angles).

Reversing the process

core concept

When you know a trig ratio (e.g. $\sin\theta = 0.6$) but not the angle, press the inverse trig function on your calculator. $\sin^{-1}(0.6)$ gives the angle whose sine is 0.6.

$$\theta = \sin^{-1}\!\left(\frac{O}{H}\right) \qquad \theta = \cos^{-1}\!\left(\frac{A}{H}\right) \qquad \theta = \tan^{-1}\!\left(\frac{O}{A}\right)$$

Converting to degrees and minutes: If your calculator gives $38.7°$, the minutes are $0.7 \times 60 = 42'$. So the answer is $38°42'$. Always round minutes to the nearest whole minute unless told otherwise.

Always check your units. Converting to consistent units before substituting is a common source of errors in assessment tasks. If you see kilometres and metres in the same problem, convert first.

What to write in your book

Inverse trig: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ — apply after you calculate the ratio.
5-step method: Label → Ratio → Evaluate → Apply inverse → Convert to D°M'.
Converting to minutes: take the decimal part and multiply by 60. e.g. $38.7° \rightarrow 38°42'$.
Both acute angles of a right triangle are complementary: $\alpha + \beta = 90°$.

Did you get this? True or false: to convert $52.4°$ to degrees and minutes, you calculate $0.4 \times 60 = 24$, giving $52°24'$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · INVERSE SINE

A right-angled triangle has opposite side 7 cm and hypotenuse 11 cm. Find the angle $\theta$ opposite the 7 cm side, in degrees and minutes.

$\sin\theta = \dfrac{O}{H} = \dfrac{7}{11} = 0.6\overline{36}$

O and H are known → use sine; calculate the ratio.

PROBLEM 2 · ANGLE OF ELEVATION

A building is 45 m tall. An observer stands 60 m from the base. Find the angle of elevation to the top of the building, in degrees and minutes.

O $= 45$ m (height), A $= 60$ m (horizontal distance). Use $\tan\theta = \dfrac{O}{A} = \dfrac{45}{60} = 0.75$

No hypotenuse needed → tangent; O is opposite the angle of elevation.

PROBLEM 3 · FINDING BOTH ANGLES

A right-angled triangle has legs 8 m and 15 m. Find both acute angles in degrees and minutes.

$\tan\alpha = \dfrac{8}{15}$; $\alpha = \tan^{-1}(0.5\overline{3}) \approx 28.07°$

Taking the angle opposite the 8 m side; use tan with O = 8, A = 15.

What to write in your book

Always label O, A, H before choosing the inverse trig function.
If O and H known: $\theta = \sin^{-1}(O/H)$; A and H: $\theta = \cos^{-1}(A/H)$; O and A: $\theta = \tan^{-1}(O/A)$.
The two acute angles in any right-angled triangle always add to 90°. Once you find one, subtract from 90° for the other.
Round minutes to the nearest whole number (e.g. $31.2' \approx 31'$).

Quick check: In a right-angled triangle, O = 5 m and H = 13 m. Which expression correctly gives the angle $\theta$?

Common errors · the 3 traps that cost marks

Trap 01

Forgetting to apply the inverse function

Writing $\sin\theta = 0.73$ and then saying $\theta = 0.73$ is a classic error. You must apply $\sin^{-1}$ to get the angle. The ratio and the angle are completely different numbers.

Trap 02

Incorrect decimal-to-minutes conversion

$39.5°$ is NOT $39°50'$. The decimal part ($0.5$) must be multiplied by 60 to get minutes: $0.5 \times 60 = 30$, so the answer is $39°30'$. A quick mental check: minutes must be less than 60.

Trap 03

Mislabelling O and A relative to the angle

O (opposite) and A (adjacent) are defined relative to the angle you are finding — not to any fixed corner of the triangle. Re-label each time you switch the angle you are working with.

What to write in your book

$\sin\theta = 0.73 \Rightarrow \theta = \sin^{-1}(0.73)$ — do not skip the inverse step.
Decimal to minutes: ×60, round to nearest whole. Minutes must be 0–59.
O and A labels change depending on which angle you are finding.
For depression problems: use the equal alternate-angle relationship to redraw as an elevation triangle.

Fill the gap: The angle $47.35°$ expressed in degrees and minutes is $47°$'. (Hint: $0.35 \times 60 = ?$)

Quick-fire practice · 8 calculations

O $= 5$ m, H $= 13$ m. Find $\theta$ in degrees and minutes.

A $= 9$ cm, H $= 15$ cm. Find $\theta$ in degrees and minutes.

O $= 7$ m, A $= 24$ m. Find $\theta$ in degrees and minutes.

A right-angled triangle has legs 5 m and 12 m. Find both acute angles in degrees and minutes.

A ramp rises 1.5 m over a horizontal distance of 9 m. Find the angle the ramp makes with the horizontal.

A 10 m ladder leans against a wall with its foot 4 m from the base. Find the angle the ladder makes with the ground.

From a lighthouse 48 m above sea level, the boat is 120 m from the base (horizontal). Find the angle of depression.

A wire is attached from the top of a 12 m pole to the ground, 5 m from the base. Find the angle the wire makes with the ground.

Odd one out: Three of these statements about inverse trig are correct. Which one is wrong?

Revisit your thinking

Earlier you thought about what to do when $\sin\theta = 0.73$. The answer: press [SHIFT][SIN] on your calculator to get $\theta = \sin^{-1}(0.73) \approx 46.89° \approx 46°53'$.

As for uniqueness — in a right-angled triangle (angles strictly between 0° and 90°), each ratio value maps to exactly one angle. So yes, there is only one possible answer in this context.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

SA 1. A right-angled triangle has a hypotenuse of 20 cm and one side of 12 cm. (a) Find the angle between the 12 cm side and the hypotenuse, in degrees and minutes. (b) Find the other acute angle. (3 marks)

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ApplyBand 43 marks

SA 2. A pilot is flying at an altitude of 1500 m. She spots a runway that is 4200 m horizontally from directly below the aircraft. (a) Draw and label the right-angled triangle. (b) Find the angle of depression from the aircraft to the runway, in degrees and minutes. (3 marks)

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AnalyseBand 54 marks

SA 3. A 15-metre telephone pole stands vertically. Safety regulations require the support wire's angle with the ground to be between 55° and 70°. (a) If the wire is anchored 9 m from the base, find the angle the wire makes with the ground (in degrees and minutes). (b) Does this satisfy the regulation? Justify. (c) Find the minimum distance from the base at which the wire can be anchored while still satisfying the regulation, to 1 decimal place. (4 marks)

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📖 Comprehensive answers (click to reveal)

Drill answers: 1: $\sin^{-1}(5/13) \approx 22°37'$ · 2: $\cos^{-1}(9/15) \approx 53°8'$ · 3: $\tan^{-1}(7/24) \approx 16°16'$ · 4: $\alpha = \tan^{-1}(5/12) \approx 22°37'$; $\beta = 67°23'$ · 5: $\tan^{-1}(1.5/9) \approx 9°28'$ · 6: $\cos^{-1}(4/10) \approx 66°25'$ · 7: $\tan^{-1}(48/120) \approx 21°48'$ · 8: $\tan^{-1}(12/5) \approx 67°23'$

SA 1 (3 marks): (a) 12 cm is adjacent to the angle; $\cos\theta = 12/20 = 0.6$; $\theta = \cos^{-1}(0.6) \approx 53.13° \approx \mathbf{53°8'}$ [2]. (b) $90° - 53°8' = \mathbf{36°52'}$ [1].

SA 2 (3 marks): (a) Right angle at the point directly below the aircraft; altitude $= 1500$ m (O); horizontal $= 4200$ m (A); angle of depression at aircraft [1]. (b) $\tan\theta = 1500/4200 = 0.3571$; $\theta = \tan^{-1}(0.3571) \approx 19.65° \approx \mathbf{19°39'}$ [2].

SA 3 (4 marks): (a) $\tan\theta = 15/9 = 1.\overline{6}$; $\theta = \tan^{-1}(5/3) \approx 59.04° \approx \mathbf{59°2'}$ [2]. (b) $55° \leq 59°2' \leq 70°$ — Yes, the angle satisfies the regulation [1]. (c) Max angle is 70°; $\tan70° = 15/d_{\min}$; $d_{\min} = 15/\tan70° \approx 15/2.747 \approx \mathbf{5.5 \text{ m}}$ [1].

Boss battle · The Angle Hunter

earn bronze · silver · gold

Five timed questions on inverse trig and degrees-and-minutes conversion. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering inverse trig questions. Pool: lesson 15.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 14 Lesson 16 · Angles of Elevation and Depression →

Module overview · Maths Standard Y11