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3.3 Right-Triangle Trigonometry and Applications

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1 Introduction & Exam Overview

1.1 AdUCET Overview, Format, and Admission Process 1.2 Test-Taking and Time-Management Strategies

2 Chapter 2: Mathematics I — Numbers, Algebra, and Functions

2.1 Number Sense, Operations, and Fraction–Decimal–Percent Conversions 2.2 Algebra, Polynomials, and Linear/Quadratic Equations 2.3 Inequalities, Absolute Value, and Systems of Equations 2.4 Radicals, Rational Expressions, and Scientific Notation 2.5 Coordinate Geometry, Lines, and Functions

3 Chapter 3: Mathematics II — Geometry, Measurement, and Trigonometry

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles 3.2 Solid Mensuration, Measurement, and Unit Conversion 3.3 Right-Triangle Trigonometry and Applications

4 Chapter 4: Mathematics III — Sequences, Proportion, and Data

4.1 Arithmetic and Geometric Sequences and Series 4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems 4.3 Descriptive Statistics and Data Interpretation 4.4 Probability: Rules, Compound Events, and Conditional Probability 4.5 Counting Principles: Permutations and Combinations

5 Chapter 5: English I — Reading Comprehension and Vocabulary

5.1 Reading Comprehension: Main Idea, Purpose, and Passage Structure 5.2 Inference, Supporting Detail, and Author's Tone 5.3 Word Roots, Prefixes, and Suffixes 5.4 Vocabulary in Context, Synonyms, Antonyms, and Confusables

6 Chapter 6: English II — Grammar, Usage, and Written Expression

6.1 Subject-Verb Agreement and Verb Tenses 6.2 Pronouns, Modifiers, and Parallel Structure 6.3 Sentence Boundaries, Clauses, and Punctuation 6.4 Sentence Correction, Redundancy, and Idiomatic Usage 6.5 Verbal Analogies and Relationship Bridges 6.6 Paragraph Organization, Coherence, and Transitions

7 Chapter 7: General Science I — Biology and Chemistry

7.1 Cell Biology, Transport, and Cellular Energetics 7.2 Genetics, Heredity, and Molecular Biology 7.3 Human Body Systems, Ecology, and Evolution 7.4 Matter, Phase Changes, and Atomic Structure 7.5 Periodic Trends, Chemical Bonding, Reactions, Acids and Bases

8 Chapter 8: General Science II — Physics and Earth Science

8.1 Mechanics: Motion, Newton's Laws, Work, and Energy 8.2 Waves, Sound, Light, Heat, Electricity, and Magnetism 8.3 Earth's Structure, Plate Tectonics, and the Rock Cycle 8.4 The Atmosphere, Weather, the Water Cycle, and Astronomy

9 Chapter 9: Aptitude and Abstract Reasoning

9.1 Figure Series and Pattern Progression 9.2 Matrix Reasoning and Figural Analogies 9.3 Spatial Reasoning: Rotation, Reflection, and Chirality 9.4 Paper Folding, Cube Nets, and Orthographic Views 9.5 Deductive Reasoning, Syllogisms, and Venn Diagrams 9.6 Number, Letter, and Alphanumeric Series 9.7 Odd-One-Out, Classification, and Logic Puzzles

3.3 Right-Triangle Trigonometry and Applications

Key Takeaways Explain This Topic AI

SOH-CAH-TOA gives sin = opposite/hypotenuse, cos = adjacent/hypotenuse and tan = opposite/adjacent; cosecant pairs with sine, secant with cosine and cotangent with tangent.

sin 30° = cos 60° = 1/2 and sin 60° = cos 30° = √3/2, so tan 30° = √3/3, tan 45° = 1 and tan 60° = √3.

The angle of elevation from A up to B equals the angle of depression from B down to A because they are alternate interior angles; both are measured from the horizontal, never from the vertical.

When two right triangles share the same unknown height, equate the shared side: elevations of 30° and 45° taken 30 m apart give h = 30/(√3 − 1) = 15(√3 + 1) ≈ 41 m.

π radians = 180°, so degrees convert to radians by multiplying by π/180; 210° = 7π/6 rad.

Last updated: August 2026

3.3 Right-Triangle Trigonometry and Applications

Right-triangle trigonometry converts an angle into a ratio of sides, which is how examiners turn an unreachable height into arithmetic. Senior-high-school trigonometry works almost entirely from the special angles , so answers stay exact: 8 3 8\sqrt{3} 8 3 ​ , 15 ( 3 + 1 ) 15(\sqrt{3}+1) 15 ( 3 ​ + 1 ) , 7 π 6 \frac{7\pi}{6} 6 7 π ​ . If a decimal is needed, use 2 ≈ 1.414 \sqrt{2}\approx 1.414 2 ​ ≈ 1.414 and 3 ≈ 1.732 \sqrt{3}\approx 1.732 3 ​ ≈ 1.732 .

1. The six ratios

Label the triangle from the angle you are working with. The hypotenuse never moves — it is always across from the right angle — but opposite and adjacent swap the moment you switch to the other acute angle. B

  • |\
  • | \
  • a | \ c right angle at C
  • | \ theta is measured at A
  • | \
  • |_____\
  • C b A

opposite = a adjacent = b hypotenuse = c

SOH-CAH-TOA : sin ⁡ θ = opp hyp , cos ⁡ θ = adj hyp , tan ⁡ θ = opp adj = sin ⁡ θ cos ⁡ θ \sin\theta = \frac{\text{opp}}{\text{hyp}},\qquad \cos\theta = \frac{\text{adj}}{\text{hyp}},\qquad \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{\sin\theta}{\cos\theta} sin θ = hyp opp ​ , cos θ = hyp adj ​ , tan θ = adj opp ​ = cos θ sin θ ​

The three reciprocals pair up in a deliberately confusing order — cosecant goes with sine, secant with cosine: csc ⁡ θ = 1 sin ⁡ θ = hyp opp , sec ⁡ θ = 1 cos ⁡ θ = hyp adj , cot ⁡ θ = 1 tan ⁡ θ = adj opp \csc\theta = \frac{1}{\sin\theta} = \frac{\text{hyp}}{\text{opp}},\qquad \sec\theta = \frac{1}{\cos\theta} = \frac{\text{hyp}}{\text{adj}},\qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\text{adj}}{\text{opp}} csc θ = sin θ 1 ​ = opp hyp ​ , sec θ = cos θ 1 ​ = adj hyp ​ , cot θ = tan θ 1 ​ = opp adj ​

Cofunction rule : sin ⁡ θ = cos ⁡ ( 90 ∘ − θ ) \sin\theta = \cos(90^\circ - \theta) sin θ = cos ( 9 0 ∘ − θ ) and tan ⁡ θ = cot ⁡ ( 90 ∘ − θ ) \tan\theta = \cot(90^\circ-\theta) tan θ = cot ( 9 0 ∘ − θ ) . That is why sin ⁡ 30 ∘ \sin 30^\circ sin 3 0 ∘ and cos ⁡ 60 ∘ \cos 60^\circ cos 6 0 ∘ are the same number.

2. The exact-value table

θ \theta θ sin ⁡ θ \sin\theta sin θ cos ⁡ θ \cos\theta cos θ tan ⁡ θ \tan\theta tan θ csc ⁡ θ \csc\theta csc θ sec ⁡ θ \sec\theta sec θ cot ⁡ θ \cot\theta cot θ

0 ∘ 0^\circ 0 ∘ 0 0 0 1 1 1 0 0 0 undefined 1 1 1 undefined

30 ∘ 30^\circ 3 0 ∘ 1 2 \frac{1}{2} 2 1 ​ 3 2 \frac{\sqrt{3}}{2} 2 3 ​ ​ 3 3 \frac{\sqrt{3}}{3} 3 3 ​ ​ 2 2 2 2 3 3 \frac{2\sqrt{3}}{3} 3 2 3 ​ ​ 3 \sqrt{3} 3 ​

45 ∘ 45^\circ 4 5 ∘ 2 2 \frac{\sqrt{2}}{2} 2 2 ​ ​ 2 2 \frac{\sqrt{2}}{2} 2 2 ​ ​ 1 1 1 2 \sqrt{2} 2 ​ 2 \sqrt{2} 2 ​ 1 1 1

60 ∘ 60^\circ 6 0 ∘ 3 2 \frac{\sqrt{3}}{2} 2 3 ​ ​ 1 2 \frac{1}{2} 2 1 ​ 3 \sqrt{3} 3 ​ 2 3 3 \frac{2\sqrt{3}}{3} 3 2 3 ​ ​ 2 2 2 3 3 \frac{\sqrt{3}}{3} 3 3 ​ ​

90 ∘ 90^\circ 9 0 ∘ 1 1 1 0 0 0 undefined 1 1 1 undefined 0 0 0

You never have to memorize this cold. Draw a 45 45 45 – 45 45 45 – 90 90 90 triangle with sides 1 , 1 , 2 1, 1, \sqrt{2} 1 , 1 , 2 ​ and a 30 30 30 – 60 60 60 – 90 90 90 triangle with sides 1 , 3 , 2 1, \sqrt{3}, 2 1 , 3 ​ , 2 , then read the ratios straight off. Note the pattern in the sine column as the angle grows: 0 2 , 1 2 , 2 2 , 3 2 , 4 2 \frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2} 2 0 ​ ​ , 2 1 ​ ​ , 2 2 ​ ​ , 2 3 ​ ​ , 2 4 ​ ​ .

3. Pythagorean identities sin ⁡ 2 θ + cos ⁡ 2 θ = 1 , 1 + tan ⁡ 2 θ = sec ⁡ 2 θ , 1 + cot ⁡ 2 θ = csc ⁡ 2 θ \sin^2\theta + \cos^2\theta = 1,\qquad 1 + \tan^2\theta = \sec^2\theta,\qquad 1 + \cot^2\theta = \csc^2\theta sin 2 θ + cos 2 θ = 1 , 1 + tan 2 θ = sec 2 θ , 1 + cot 2 θ = csc 2 θ

The second and third are just the first divided through by cos ⁡ 2 θ \cos^2\theta cos 2 θ and sin ⁡ 2 θ \sin^2\theta sin 2 θ . Use them to jump between ratios without finding the angle: if θ \theta θ is acute and sin ⁡ θ = 3 5 \sin\theta = \frac{3}{5} sin θ = 5 3 ​ , then cos ⁡ 2 θ = 1 − 9 25 = 16 25 \cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25} cos 2 θ = 1 − 25 9 ​ = 25 16 ​ , so cos ⁡ θ = 4 5 \cos\theta = \frac{4}{5} cos θ = 5 4 ​ (positive, because θ \theta θ is acute) and tan ⁡ θ = 3 4 \tan\theta = \frac{3}{4} tan θ = 4 3 ​ . The faster route is to recognize the 3 3 3 - 4 4 4 - 5 5 5 triangle directly.

4. Finding an unknown side or angle

  • You know You want Set up
  • angle + adjacent opposite opp = adj × tan ⁡ θ \text{opp} = \text{adj}\times\tan\theta opp = adj × tan θ
  • angle + opposite adjacent adj = opp tan ⁡ θ \text{adj} = \dfrac{\text{opp}}{\tan\theta} adj = tan θ opp ​
  • angle + hypotenuse opposite opp = hyp × sin ⁡ θ \text{opp} = \text{hyp}\times\sin\theta opp = hyp × sin θ
  • angle + opposite hypotenuse hyp = opp sin ⁡ θ \text{hyp} = \dfrac{\text{opp}}{\sin\theta} hyp = sin θ opp ​
  • two sides the angle match the ratio to the table

When the unknown sits in the denominator, cross-multiply rather than guess: sin ⁡ 30 ∘ = 12 x \sin 30^\circ = \frac{12}{x} sin 3 0 ∘ = x 12 ​ means x = 12 sin ⁡ 30 ∘ = 12 1 / 2 = 24 x = \frac{12}{\sin 30^\circ} = \frac{12}{1/2} = 24 x = s i n 3 0 ∘ 12 ​ = 1/2 12 ​ = 24 , not 6 6 6 . Because only special angles appear, "finding the angle" means recognizing a value: tan ⁡ θ = 1 ⇒ θ = 45 ∘ \tan\theta = 1 \Rightarrow \theta = 45^\circ tan θ = 1 ⇒ θ = 4 5 ∘ ; tan ⁡ θ = 3 ⇒ θ = 60 ∘ \tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ tan θ = 3 ​ ⇒ θ = 6 0 ∘ ; sin ⁡ θ = 1 2 ⇒ θ = 30 ∘ \sin\theta = \frac{1}{2} \Rightarrow \theta = 30^\circ sin θ = 2 1 ​ ⇒ θ = 3 0 ∘ .

5. Angles of elevation and depression

Angle of elevation : measured upward from the observer's horizontal line of sight to the object.

Angle of depression : measured downward from the observer's horizontal line of sight. observer A ======================== horizontal through A

  • | \
  • | \ <- angle of depression, measured at A
  • h | \ between the horizontal and the line of sight
  • | \
  • +-----B ===================== ground: horizontal through B
  • x <- angle of elevation, measured at B
  • (equal to the depression angle at A)

Since the two horizontal lines are parallel, the angle of elevation from the ground point up to the observer equals the angle of depression from the observer down to that point — they are alternate interior angles . Depression angles are therefore usable inside the triangle without any extra step, provided you place them at the bottom vertex.

6. Worked example 1 — flagpole and shadow

Problem. A flagpole on level ground casts a 24 24 24 m shadow at the moment the sun's angle of elevation is 30 ∘ 30^\circ 3 0 ∘ . A shorter pole beside it casts a 9 9 9 m shadow at the same instant. Find both heights.

The shadow is the adjacent side and the pole is the opposite side, so use tangent: tan ⁡ 30 ∘ = h 24 \tan 30^\circ = \frac{h}{24} tan 3 0 ∘ = 24 h ​ .

h = 24 tan ⁡ 30 ∘ = 24 3 = 24 3 3 = 8 3 ≈ 13.86 h = 24\tan 30^\circ = \frac{24}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \approx 13.86 h = 24 tan 3 0 ∘ = 3 ​ 24 ​ = 3 24 3 ​ ​ = 8 3 ​ ≈ 13.86 m.

For the shorter pole, h ′ = 9 tan ⁡ 30 ∘ = 9 3 = 3 3 ≈ 5.20 h' = 9\tan 30^\circ = \frac{9}{\sqrt{3}} = 3\sqrt{3} \approx 5.20 h ′ = 9 tan 3 0 ∘ = 3 ​ 9 ​ = 3 3 ​ ≈ 5.20 m.

Sanity check by similarity: at one instant the sun's angle is the same for both, so heights are proportional to shadows, and 8 3 3 3 = 8 3 = 24 9 \frac{8\sqrt{3}}{3\sqrt{3}} = \frac{8}{3} = \frac{24}{9} 3 3 ​ 8 3 ​ ​ = 3 8 ​ = 9 24 ​ .

7. Worked example 2 — one height shared by two triangles

Problem. Standing on level ground along Taft Avenue, Maricel measures the angle of elevation to the top of a condominium as 30 ∘ 30^\circ 3 0 ∘ . She walks 30 30 30 m straight toward the building and the angle of elevation becomes 45 ∘ 45^\circ 4 5 ∘ . How tall is the building above her eye level, and how tall is it in total if her eyes are 1.5 1.5 1.5 m above the ground?

Let h h h be the height above eye level and x x x the horizontal distance from the nearer position to the base. The unknown h h h is the shared side of both right triangles.

Near triangle: tan ⁡ 45 ∘ = h x ⇒ 1 = h x ⇒ x = h \tan 45^\circ = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h tan 4 5 ∘ = x h ​ ⇒ 1 = x h ​ ⇒ x = h .

Far triangle: tan ⁡ 30 ∘ = h x + 30 \tan 30^\circ = \frac{h}{x + 30} tan 3 0 ∘ = x + 30 h ​ , and substituting x = h x = h x = h gives 1 3 = h h + 30 \frac{1}{\sqrt{3}} = \frac{h}{h+30} 3 ​ 1 ​ = h + 30 h ​ .

Cross-multiply: h + 30 = h 3 ⇒ 30 = h ( 3 − 1 ) ⇒ h = 30 3 − 1 h + 30 = h\sqrt{3} \Rightarrow 30 = h(\sqrt{3}-1) \Rightarrow h = \frac{30}{\sqrt{3}-1} h + 30 = h 3 ​ ⇒ 30 = h ( 3 ​ − 1 ) ⇒ h = 3 ​ − 1 30 ​ .

Rationalize: h = 30 ( 3 + 1 ) ( 3 − 1 ) ( 3 + 1 ) = 30 ( 3 + 1 ) 2 = 15 ( 3 + 1 ) ≈ 15 ( 2.732 ) = 40.98 h = \frac{30(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{30(\sqrt{3}+1)}{2} = 15(\sqrt{3}+1) \approx 15(2.732) = 40.98 h = ( 3 ​ − 1 ) ( 3 ​ + 1 ) 30 ( 3 ​ + 1 ) ​ = 2 30 ( 3 ​ + 1 ) ​ = 15 ( 3 ​ + 1 ) ≈ 15 ( 2.732 ) = 40.98 m.

Total height ≈ 40.98 + 1.5 = 42.5 \approx 40.98 + 1.5 = 42.5 ≈ 40.98 + 1.5 = 42.5 m. Forgetting to add eye height is the built-in trap.

8. Bearings and simple navigation

Two notations appear:

Quadrant bearing — N 30 ∘ E N\,30^\circ\,E N 3 0 ∘ E means start at north and rotate 30 ∘ 30^\circ 3 0 ∘ toward east.

True (azimuth) bearing — 030 ∘ 030^\circ 03 0 ∘ , always three digits, measured clockwise from north.

So N 30 ∘ E N\,30^\circ\,E N 3 0 ∘ E and 030 ∘ 030^\circ 03 0 ∘ describe the same heading, while S 40 ∘ W S\,40^\circ\,W S 4 0 ∘ W equals 220 ∘ 220^\circ 22 0 ∘ . To resolve a leg of a journey, draw the north–south line and use the angle at the starting point : a boat sailing 40 40 40 km on bearing N 30 ∘ E N\,30^\circ\,E N 3 0 ∘ E travels east = 40 sin ⁡ 30 ∘ = 20 km , north = 40 cos ⁡ 30 ∘ = 20 3 ≈ 34.6 km . \text{east} = 40\sin 30^\circ = 20\ \text{km},\qquad \text{north} = 40\cos 30^\circ = 20\sqrt{3} \approx 34.6\ \text{km}. east = 40 sin 3 0 ∘ = 20 km , north = 40 cos 3 0 ∘ = 20 3 ​ ≈ 34.6 km .

Notice that sine takes the east–west component here because the angle is measured from north , the reverse of the usual habit.

9. Degrees and radians π rad = 180 ∘ ⟹ radians = degrees × π 180 , degrees = radians × 180 π \pi\ \text{rad} = 180^\circ \quad\Longrightarrow\quad \text{radians} = \text{degrees}\times\frac{\pi}{180},\qquad \text{degrees} = \text{radians}\times\frac{180}{\pi} π rad = 18 0 ∘ ⟹ radians = degrees × 180 π ​ , degrees = radians × π 180 ​

Degrees 30 30 30 45 45 45 60 60 60 90 90 90 120 120 120 135 135 135 180 180 180 210 210 210 270 270 270 360 360 360

Radians π 6 \frac{\pi}{6} 6 π ​ π 4 \frac{\pi}{4} 4 π ​ π 3 \frac{\pi}{3} 3 π ​ π 2 \frac{\pi}{2} 2 π ​ 2 π 3 \frac{2\pi}{3} 3 2 π ​ 3 π 4 \frac{3\pi}{4} 4 3 π ​ π \pi π 7 π 6 \frac{7\pi}{6} 6 7 π ​ 3 π 2 \frac{3\pi}{2} 2 3 π ​ 2 π 2\pi 2 π

10. Common traps

Reversing opposite and adjacent after switching to the other acute angle.

Measuring an angle of depression from the vertical tower instead of from the horizontal.

Using sine when the two known quantities are the two legs, where tangent is required.

Multiplying by the ratio when the unknown is a hypotenuse in the denominator and you should divide.

Omitting the observer's eye height or the height of a cliff base.

Treating N 30 ∘ E N\,30^\circ\,E N 3 0 ∘ E as 30 ∘ 30^\circ 3 0 ∘ from east.

Leaving an answer in degrees when the choices are in radians.