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.

55–60 min MS-M2 — MEDIUM 3 MC 3 SA Lesson 15 of 22 Free
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Think First

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 Trigonometric Functions — This Lesson

$\theta = \sin^{-1}\!\left(\dfrac{O}{H}\right)$
Inverse sine — use when opposite (O) and hypotenuse (H) are known Calculator: [SHIFT] [SIN] or [SIN⁻¹]
$\theta = \cos^{-1}\!\left(\dfrac{A}{H}\right)$
Inverse cosine — use when adjacent (A) and hypotenuse (H) are known
$\theta = \tan^{-1}\!\left(\dfrac{O}{A}\right)$
Inverse tangent — use when opposite (O) and adjacent (A) are known (no hypotenuse needed)
$x° \rightarrow x° \; y'$
Converting to degrees and minutes: take the decimal part and multiply by 60 e.g. $38.7° \rightarrow 38°42'$ (since $0.7 \times 60 = 42$)
FINDING AN UNKNOWN ANGLE θ=? 7 12 (O) (A) O and A known → use TOA: θ = tan⁻¹(7/12) θ ≈ 30.26° DECIMAL DEGREES → D° M' Start: 30.26° whole part = 30° — keep this Decimal: 0.26 × 60 multiply decimal part by 60 = 30° 15.6' ≈ 30° 16' Round minutes to nearest whole minute

🧠 Know

  • $\theta = \sin^{-1}$, $\cos^{-1}$, 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

  • 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

  • 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
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Key Terms

Inverse trigonometric function The "reverse" of a trig function: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ take a ratio value and return the angle
Degrees and minutes Angles expressed as whole degrees (°) plus minutes (') where 1° = 60'; convert decimal part by multiplying by 60
Angle of elevation The angle measured upward from the horizontal line of sight to an object above; always creates a right-angled triangle
Angle of depression The 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)

Misconceptions to Fix

Wrong: Rounding 4.5 down gives 4.

Right: Standard rounding rounds 4.5 up to 5; rounding to the nearest even number (banker's rounding) is different.

Key Point

Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.

Key Terms
FormulaA rule showing the relationship between variables using symbols.
SubstitutionReplacing variables with their known values in an equation.
Unit ConversionChanging a measurement from one unit to another.
CapacityThe amount of liquid a container can hold, measured in litres or millilitres.
PerimeterThe total distance around the outside of a shape.
AreaThe amount of space inside a two-dimensional shape.

Reversing the Process

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.

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'.
Worked Example 1 Inverse Sine

Problem

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.

Solution

1 $\sin\theta = \dfrac{O}{H} = \dfrac{7}{11} = 0.6\overline{36}$ O and H are known → use sine; calculate the ratio
Worked Example 2 Angle of Elevation

Problem

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.

Solution

1 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
Worked Example 3 Finding Both Angles

Problem

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

Solution

1 $\tan\alpha = \dfrac{8}{15}$; $\alpha = \tan^{-1}(0.5\overline{3}) \approx 28.07°$ Taking the angle opposite the 8 m side; use tan with O=8, A=15
Practice

Practice Questions

Express all angle answers in degrees and minutes unless told otherwise.

Section A — Finding a Single Angle

  1. O $= 5$ m, H $= 13$ m. Find $\theta$.
  2. A $= 9$ cm, H $= 15$ cm. Find $\theta$.
  3. O $= 7$ m, A $= 24$ m. Find $\theta$.
  4. O $= 4.2$ cm, H $= 9.8$ cm. Find $\theta$.
  5. A $= 8.5$ m, H $= 14.0$ m. Find $\theta$.

Section B — Finding Both Angles

  1. A right-angled triangle has legs 5 m and 12 m. Find both acute angles.
  2. A right-angled triangle has hypotenuse 25 cm and one leg 7 cm. Find both acute angles.

Section C — Practical Applications

  1. A ramp rises 1.5 m over a horizontal distance of 9 m. Find the angle the ramp makes with the horizontal.
  2. A ship travels 40 km north then 30 km east. Find the angle its path deviates from north.
  3. From a lighthouse 48 m above sea level, the angle of depression to a boat is to be found. The boat is 120 m from the base of the lighthouse (horizontal). Find the angle of depression.
  4. A ski slope descends 280 m vertically over a horizontal distance of 650 m. Find the angle of inclination of the slope.
  5. 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.
  6. From the top of a building 80 m tall, two cars are observed. Car A is at an angle of depression of 32° and Car B is at an angle of depression of 18°. Find the horizontal distance to each car.

Q1

$\sin^{-1}(5/13) \approx 22.62° \approx \mathbf{22°37'}$

Q2

$\cos^{-1}(9/15) = \cos^{-1}(0.6) \approx 53.13° \approx \mathbf{53°8'}$

Q3

$\tan^{-1}(7/24) \approx 16.26° \approx \mathbf{16°16'}$

Q4

$\sin^{-1}(4.2/9.8) \approx \sin^{-1}(0.4286) \approx 25.38° \approx \mathbf{25°23'}$

Q5

$\cos^{-1}(8.5/14) \approx \cos^{-1}(0.6071) \approx 52.61° \approx \mathbf{52°37'}$

Q6

$\alpha = \tan^{-1}(5/12) \approx 22°37'$; $\beta = 90° - 22°37' = \mathbf{67°23'}$

Q7

One leg $= 7$, H $= 25$; $\alpha = \sin^{-1}(7/25) \approx 16°16'$; $\beta = 90° - 16°16' = \mathbf{73°44'}$

Q8

$\tan^{-1}(1.5/9) \approx \tan^{-1}(0.1\overline{6}) \approx 9.46° \approx \mathbf{9°28'}$

Q9

$\tan^{-1}(30/40) = \tan^{-1}(0.75) \approx 36.87° \approx \mathbf{36°52'}$ east of north

Q10

$\tan^{-1}(48/120) = \tan^{-1}(0.4) \approx 21.80° \approx \mathbf{21°48'}$

Q11

$\tan^{-1}(280/650) \approx \tan^{-1}(0.4308) \approx 23.32° \approx \mathbf{23°19'}$

Q12

$\tan^{-1}(12/5) \approx 67.38° \approx \mathbf{67°23'}$

Q13

Car A: $d = 80/\tan32° \approx \mathbf{128.0 \text{ m}}$; Car B: $d = 80/\tan18° \approx \mathbf{246.2 \text{ m}}$

Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Multiple Choice

1 In a right-angled triangle, the side opposite angle $\theta$ is 9 m and the hypotenuse is 14 m. Angle $\theta$, in degrees and minutes, is closest to:

A   39°45'
B   40°1'
C   49°59'
D   50°15'

? Regarding this topic, 1 In a right-angled triangle, the side opposite angle $\theta$ is 9 m and the hypotenuse is 14 m. Angle $\theta$, in degrees and minutes, is closest to:

A     39°45'
B     40°1'
C     49°59'
D     50°15'
B - Correct!
B — $\sin^{-1}(9/14) = \sin^{-1}(0.6429) \approx 40.01° \approx 40°1'$.

2 A 10-metre ladder leans against a vertical wall. The foot of the ladder is 4.8 m from the base of the wall. The angle the ladder makes with the ground, in degrees and minutes, is closest to:

A   28°41'
B   28°48'
C   61°12'
D   61°19'

? Regarding this topic, 2 A 10-metre ladder leans against a vertical wall. The foot of the ladder is 4.8 m from the base of the wall. The angle the ladder makes with the ground, in degrees and minutes, is closest to:

A     28°41'
B     28°48'
C     61°12'
D     61°19'
D - Correct!
D — $\cos^{-1}(4.8/10) = \cos^{-1}(0.48) \approx 61.315... ° \approx 61°19'$. (The angle with the ground, not the wall.)

3 A vertical cliff is 75 m high. From the base, the angle of elevation to the top of a tree on top of the cliff is 42°. The horizontal distance from the base of the cliff to the observer is 85 m. The angle of elevation to the bottom of the cliff is:

A   38°20'
B   41°26'
C   48°34'
D   41°22'

? Regarding this topic, 3 A vertical cliff is 75 m high. From the base, the angle of elevation to the top of a tree on top of the cliff is 42°. The horizontal distance from the base of the cliff to the observer is 85 m. The angle of elevation to the bottom of the cliff is:

A     38°20'
B     41°26'
C     48°34'
D     41°22'
D - Correct!
D — $\tan^{-1}(75/85) = \tan^{-1}(0.8824) \approx 41.37° \approx 41°22'$.

Short Answer

01

SA 4 3 marks 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.  (2 marks)

(b) Find the other acute angle.  (1 mark)

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(a)

12 cm is adjacent to the angle (angle between it and H); $\cos\theta = 12/20 = 0.6$; $\theta = \cos^{-1}(0.6) \approx 53.13° \approx \mathbf{53°8'}$

(b)

$90° - 53°8' = \mathbf{36°52'}$

02

SA 5 3 marks 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.  (1 mark)

(b) Find the angle of depression from the aircraft to the runway, in degrees and minutes.  (2 marks)

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(a)

Right angle at the point directly below the aircraft; altitude $= 1500$ m (O); horizontal distance $= 4200$ m (A); angle of depression at aircraft position

(b)

$\tan\theta = 1500/4200 = 0.3571$; $\theta = \tan^{-1}(0.3571) \approx 19.65° \approx \mathbf{19°39'}$

03

SA 6 4 marks A 15-metre telephone pole stands vertically. A support wire is attached from the top of the pole to a point on the ground. Safety regulations require the angle between the wire and the ground to be between 55° and 70°.

(a) If the wire is anchored 9 m from the base of the pole, find the angle the wire makes with the ground (in degrees and minutes).  (2 marks)

(b) Does this satisfy the safety regulation? Justify your answer.  (1 mark)

(c) Find the minimum distance from the base of the pole at which the wire can be anchored while still satisfying the regulation. Give your answer to 1 decimal place.  (1 mark)

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(a)

$\tan\theta = 15/9 = 1.\overline{6}$; $\theta = \tan^{-1}(5/3) \approx 59.04° \approx \mathbf{59°2'}$

(b)

$55° \leq 59°2' \leq 70°$ — Yes, the angle satisfies the regulation.

(c)

Max angle is 70°; $\tan70° = 15/d_{\min}$; $d_{\min} = 15/\tan70° \approx 15/2.747 \approx \mathbf{5.5 \text{ m}}$

Consolidation Game

Right-Angled Trigonometry: Finding Unknown Angles