Study guide
2.2 Trigonometry & Solid Mensuration
PRC Electronics Engineering Licensure Examination Study Guide Exam
Track your progress as you read
Finished sections get a check mark.
1 Chapter 1: Introduction & Board Exam Strategy
1.1 PRC ECE Licensure Exam Overview & Regulation 1.2 Table of Specifications & Exam Strategy
2 Chapter 2: Mathematics Foundations
2.1 College & Advanced Algebra 2.2 Trigonometry & Solid Mensuration 2.3 Analytic Geometry & Differential Calculus 2.4 Integral Calculus & Applications
3 Chapter 3: Advanced Mathematics for ECE
3.1 Differential Equations & Circuit Transients 3.2 Complex Numbers & Laplace Transforms 3.3 Fourier Series, Fourier Transforms & Z-Transforms 3.4 Vector Analysis, Matrices & Probability
4 Chapter 4: GEAS - Engineering Sciences & Economics
4.1 Engineering Physics & Materials Chemistry 4.2 Engineering Mechanics & Strength of Materials 4.3 Engineering Economics & Financial Analysis 4.4 Computer Fundamentals & C/C++ Programming
5 Chapter 5: GEAS - Management, Materials, Environment & CAD
5.1 Engineering Management & Technopreneurship 101 5.2 Materials Science & Engineering 5.3 Environment Science & Engineering 5.4 Computer-Aided Drawing (CAD) & Engineering Graphics
6 Chapter 6: GEAS - ECE Laws, Ethics & Contracts
6.1 Republic Act No. 9292 (Electronics Engineering Law of 2004) 6.2 ECE Code of Ethics & Professional Conduct 6.3 Contracts, Specifications & National Building Code (PD 1096)
7 Chapter 7: Electronics Engineering - Circuits & Semiconductor Diodes
7.1 DC & AC Circuit Analysis & Network Theorems 7.2 Electromagnetics & Passive Components 7.3 Semiconductor Physics & Diode Circuits 7.4 Power Supply Design & Voltage Regulators
8 Chapter 8: Electronics Engineering - Transistors & Analog Circuits
8.1 BJT Operation, Biasing & Small-Signal Amplifiers 8.2 FET Characteristics, Biasing & Amplifier Circuits 8.3 Multistage Amplifiers, Frequency Response & Power Amplifiers 8.4 Operational Amplifiers & Active Filter Design
9 Chapter 9: Electronics Engineering - Digital Electronics & Control Systems
9.1 Feedback Topologies & Oscillator Circuits 9.2 Boolean Algebra, Logic Gates & Combinational Circuits 9.3 Sequential Logic, Flip-Flops & Microprocessor Architecture 9.4 Industrial Power Electronics & Control System Theory
10 Chapter 10: EST - Communications Fundamentals & Transmission Lines
10.1 Signal Fundamentals, Decibel Calculations & Noise Analysis 10.2 Amplitude Modulation (AM) Principles & Receivers 10.3 Frequency Modulation (FM), Phase Modulation & Discriminators 10.4 Transmission Line Parameters, SWR & Smith Chart
11 Chapter 11: EST - Digital Communications, Wireless & Optical Systems
11.1 Pulse & Digital Modulation Systems 11.2 Data Communications, OSI Model & Protocols 11.3 Antenna Principles & Electromagnetic Wave Propagation 11.4 Optical Fiber, Microwave, Satellite & Philippine NTC Regulations
2.2 Trigonometry & Solid Mensuration
Key Takeaways Explain This Topic AI
Apply plane trigonometric identities and Euler's formula to perform sinusoidal AC phasor analysis and impedance calculations.
Utilize the Law of Sines and Law of Cosines to solve non-right vector triangles in multi-phase power systems and antenna array design.
Solve right spherical triangles using Napier's Rules for satellite azimuth/elevation and great-circle radio propagation paths.
Calculate surface areas and volumes of complex geometric solids and solids of revolution using Pappus-Guldinus theorems.
Last updated: July 2026
2.2 Trigonometry & Solid Mensuration
Trigonometric relationships and geometric mensuration are fundamental to understanding wave propagation, AC circuit phasors, antenna geometry, and physical component design in electronics engineering.
- 1. Plane Trigonometry & AC Phasor Analysis
- Fundamental Trigonometric Identities
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 \quad | \quad 1 + \tan^2 \theta = \sec^2 \theta \quad | \quad 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
Angle Sum & Difference Identities : sin ( α ± β ) = sin α cos β ± cos α sin β \sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta sin ( α ± β ) = sin α cos β ± cos α sin β cos ( α ± β ) = cos α cos β ∓ sin α sin β \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta cos ( α ± β ) = cos α cos β ∓ sin α sin β
Double-Angle Identities : sin ( 2 θ ) = 2 sin θ cos θ \sin(2\theta) = 2\sin\theta\cos\theta sin ( 2 θ ) = 2 sin θ cos θ cos ( 2 θ ) = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ \cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta cos ( 2 θ ) = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ
Euler's Identity & Phasor Representation
In AC electrical engineering, sinusoidal voltages and currents are modeled as rotating complex phasors using Euler's formula: ≠ j θ = cos θ + j sin θ \ne^{j\theta} = \cos\theta + j\sin\theta = j θ = cos θ + j sin θ
An AC voltage waveform v ( t ) = V m cos ( ω t + ϕ ) v(t) = V_m \cos(\omega t + \phi) v ( t ) = V m cos ( ω t + ϕ ) is represented in polar phasor form as V = V m ∠ ϕ = V m ( cos ϕ + j sin ϕ ) \mathbf{V} = V_m \angle \phi = V_m (\cos\phi + j\sin\phi) V = V m ∠ ϕ = V m ( cos ϕ + j sin ϕ ) .
2. Law of Sines & Law of Cosines
For any oblique plane triangle with sides a , b , c a, b, c a , b , c and opposite angles A , B , C A, B, C A , B , C :
Law of Sines : a sin A = b sin B = c sin C \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} sin A a = sin B b = sin C c
- Law of Cosines : c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab \cos C c 2 = a 2 + b 2 − 2 ab cos C
- Worked ECE Board Example 1: AC Voltage Vector Addition
Problem: Two AC voltage sources connected in series produce voltages v 1 ( t ) = 80 cos ( ω t ) V v_1(t) = 80 \cos(\omega t)\text{ V} v 1 ( t ) = 80 cos ( ω t ) V and v 2 ( t ) = 60 cos ( ω t + 60 ∘ ) V v_2(t) = 60 \cos(\omega t + 60^\circ)\text{ V} v 2 ( t ) = 60 cos ( ω t + 6 0 ∘ ) V . Find the magnitude of the total peak voltage V T V_T V T and its phase angle θ T \theta_T θ T .
Solution:
Step 1: Represent as vectors V 1 = 80 ∠ 0 ∘ \mathbf{V}_1 = 80 \angle 0^\circ V 1 = 80∠ 0 ∘ and V 2 = 60 ∠ 60 ∘ \mathbf{V}_2 = 60 \angle 60^\circ V 2 = 60∠6 0 ∘ .
Step 2: In the vector addition triangle, the angle between V 1 \mathbf{V}_1 V 1 and V 2 \mathbf{V}_2 V 2 is 180 ∘ − 60 ∘ = 120 ∘ 180^\circ - 60^\circ = 120^\circ 18 0 ∘ − 6 0 ∘ = 12 0 ∘ . Apply the Law of Cosines to find resultant magnitude V T V_T V T : V T 2 = 80 2 + 60 2 − 2 ( 80 ) ( 60 ) cos ( 120 ∘ ) V_T^2 = 80^2 + 60^2 - 2(80)(60)\cos(120^\circ) V T 2 = 8 0 2 + 6 0 2 − 2 ( 80 ) ( 60 ) cos ( 12 0 ∘ ) V T 2 = 6400 + 3600 − 9600 ( − 0.5 ) = 10000 + 4800 = 14800 V_T^2 = 6400 + 3600 - 9600(-0.5) = 10000 + 4800 = 14800 V T 2 = 6400 + 3600 − 9600 ( − 0.5 ) = 10000 + 4800 = 14800 V T = 14800 = 121.66 V V_T = \sqrt{14800} = 121.66\text{ V} V T = 14800 = 121.66 V
Step 3: Apply the Law of Sines to find phase angle θ T \theta_T θ T : 60 sin θ T = 121.66 sin ( 120 ∘ ) ⟹ sin θ T = 60 × 0.866025 121.66 = 51.9615 121.66 = 0.4271 \frac{60}{\sin\theta_T} = \frac{121.66}{\sin(120^\circ)} \implies \sin\theta_T = \frac{60 \times 0.866025}{121.66} = \frac{51.9615}{121.66} = 0.4271 sin θ T 60 = sin ( 12 0 ∘ ) 121.66 ⟹ sin θ T = 121.66 60 × 0.866025 = 121.66 51.9615 = 0.4271 θ T = arcsin ( 0.4271 ) = 25.28 ∘ \theta_T = \arcsin(0.4271) = 25.28^\circ θ T = arcsin ( 0.4271 ) = 25.2 8 ∘ Total voltage vector: V T = 121.66 ∠ 25.28 ∘ V \mathbf{V}_T = 121.66 \angle 25.28^\circ\text{ V} V T = 121.66∠25.2 8 ∘ V .
3. Spherical Trigonometric Relations & Napier's Rules
Spherical trigonometry deals with triangles drawn on the surface of a sphere, where the sides a , b , c a, b, c a , b , c are measured as angular arc lengths (in degrees or radians). This is crucial in ECE board problems involving satellite communications, GPS positioning, and antenna beam direction finding.
Napier's Circle for Right Spherical Triangles
For a right spherical triangle (where C = 90 ∘ C = 90^\circ C = 9 0 ∘ ), eliminate right angle C C C and arrange the remaining 5 parts in circular order: a a a , b b b , co- A \text{co-}A co- A ( 90 ∘ − A 90^\circ - A 9 0 ∘ − A ), co- c \text{co-}c co- c ( 90 ∘ − c 90^\circ - c 9 0 ∘ − c ), and co- B \text{co-}B co- B ( 90 ∘ − B 90^\circ - B 9 0 ∘ − B ).
Napier's Two Rules
Rule 1 (Sine-TaN Rule) : The sine of any middle part is equal to the product of the tangents of the two adjacent parts: sin ( middle ) = tan ( adjacent 1 ) ⋅ tan ( adjacent 2 ) \sin(\text{middle}) = \tan(\text{adjacent}_1) \cdot \tan(\text{adjacent}_2) sin ( middle ) = tan ( adjacent 1 ) ⋅ tan ( adjacent 2 )
Rule 2 (Sine-CoS Rule) : The sine of any middle part is equal to the product of the cosines of the two opposite parts: sin ( middle ) = cos ( opposite 1 ) ⋅ cos ( opposite 2 ) \sin(\text{middle}) = \cos(\text{opposite}_1) \cdot \cos(\text{opposite}_2) sin ( middle ) = cos ( opposite 1 ) ⋅ cos ( opposite 2 )
Worked ECE Board Example 2: Satellite Earth Station Great-Circle Distance
Problem: A satellite tracking system models a right spherical triangle on Earth's surface with side a = 30 ∘ a = 30^\circ a = 3 0 ∘ and side b = 45 ∘ b = 45^\circ b = 4 5 ∘ (where angle C = 90 ∘ C = 90^\circ C = 9 0 ∘ ). Calculate the great-circle arc length c c c in degrees and in kilometers (taking Earth's radius R = 6 , 371 km R = 6,371\text{ km} R = 6 , 371 km ).
Solution:
Step 1: Select middle part co- c = 90 ∘ − c \text{co-}c = 90^\circ - c co- c = 9 0 ∘ − c . Its opposite parts are a a a and b b b .
Step 2: Apply Napier's Rule 2 (Sine-CoS Rule): sin ( co- c ) = cos ( a ) ⋅ cos ( b ) ⟹ sin ( 90 ∘ − c ) = cos a cos b \sin(\text{co-}c) = \cos(a) \cdot \cos(b) \implies \sin(90^\circ - c) = \cos a \cos b sin ( co- c ) = cos ( a ) ⋅ cos ( b ) ⟹ sin ( 9 0 ∘ − c ) = cos a cos b cos c = cos ( 30 ∘ ) ⋅ cos ( 45 ∘ ) = ( 0.866025 ) ⋅ ( 0.707107 ) = 0.612372 \cos c = \cos(30^\circ) \cdot \cos(45^\circ) = (0.866025) \cdot (0.707107) = 0.612372 cos c = cos ( 3 0 ∘ ) ⋅ cos ( 4 5 ∘ ) = ( 0.866025 ) ⋅ ( 0.707107 ) = 0.612372 c = arccos ( 0.612372 ) = 52.24 ∘ c = \arccos(0.612372) = 52.24^\circ c = arccos ( 0.612372 ) = 52.2 4 ∘
Step 3: Convert angular distance c c c to linear distance along Earth's surface: c rad = 52.239 ∘ × π 180 ∘ = 0.91174 rad c_{\text{rad}} = 52.239^\circ \times \frac{\pi}{180^\circ} = 0.91174\text{ rad} c rad = 52.23 9 ∘ × 18 0 ∘ π = 0.91174 rad D = R ⋅ c rad = 6371 km × 0.91174 = 5 , 808.7 km D = R \cdot c_{\text{rad}} = 6371\text{ km} \times 0.91174 = 5,808.7\text{ km} D = R ⋅ c rad = 6371 km × 0.91174 = 5 , 808.7 km
- 4. Solid Mensuration & Theorems of Pappus-Guldinus
- Standard Solid Figures
Right Circular Cone : V = 1 3 π r 2 h V = \frac{1}{3}\pi r^2 h V = 3 1 π r 2 h , Lateral Area A L = π r s A_L = \pi r s A L = π rs (where slant height s = r 2 + h 2 s = \sqrt{r^2 + h^2} s = r 2 + h 2 ).
Frustum of a Cone : V = 1 3 π h ( R 2 + r 2 + R r ) V = \frac{1}{3}\pi h (R^2 + r^2 + R r) V = 3 1 πh ( R 2 + r 2 + R r ) .
Sphere : V = 4 3 π r 3 V = \frac{4}{3}\pi r^3 V = 3 4 π r 3 , Surface Area A = 4 π r 2 A = 4\pi r^2 A = 4 π r 2 .
Theorems of Pappus-Guldinus for Solids of Revolution
First Theorem (Surface Area) : The area of the surface generated by revolving a plane curve about a non-intersecting axis in its plane equals the product of the length of the curve L L L and the distance traveled by its centroid y ˉ \bar{y} y ˉ : A = 2 π y ˉ L A = 2\pi \bar{y} L A = 2 π y ˉ L
Second Theorem (Volume) : The volume of the solid generated by revolving a plane area A plane A_{\text{plane}} A plane about a non-intersecting axis in its plane equals the product of the area and the distance traveled by its centroid y ˉ \bar{y} y ˉ : V = 2 π y ˉ A plane V = 2\pi \bar{y} A_{\text{plane}} V = 2 π y ˉ A plane
Worked ECE Board Example 3: Toroidal Ferrite Core Volume & Surface Area
Problem: A toroidal inductor core is manufactured by revolving a circular cross-section of radius r = 2 cm r = 2\text{ cm} r = 2 cm around an axis located at a distance R = 10 cm R = 10\text{ cm} R = 10 cm from the center of the circle. Calculate the total volume and total surface area of the ferrite core.
Solution:
Step 1: Compute plane cross-sectional area and perimeter: A plane = π r 2 = π ( 2 ) 2 = 4 π cm 2 A_{\text{plane}} = \pi r^2 = \pi (2)^2 = 4\pi\text{ cm}^2 A plane = π r 2 = π ( 2 ) 2 = 4 π cm 2 L = 2 π r = 2 π ( 2 ) = 4 π cm L = 2\pi r = 2\pi (2) = 4\pi\text{ cm} L = 2 π r = 2 π ( 2 ) = 4 π cm
Step 2: Centroid distance to axis of rotation is y ˉ = R = 10 cm \bar{y} = R = 10\text{ cm} y ˉ = R = 10 cm .
Step 3: Apply Pappus's Second Theorem for Volume: V = 2 π y ˉ A plane = 2 π ( 10 ) ( 4 π ) = 80 π 2 ≈ 789.57 cm 3 V = 2\pi \bar{y} A_{\text{plane}} = 2\pi (10)(4\pi) = 80\pi^2 \approx 789.57\text{ cm}^3 V = 2 π y ˉ A plane = 2 π ( 10 ) ( 4 π ) = 80 π 2 ≈ 789.57 cm 3
Step 4: Apply Pappus's First Theorem for Surface Area: A = 2 π y ˉ L = 2 π ( 10 ) ( 4 π ) = 80 π 2 ≈ 789.57 cm 2 A = 2\pi \bar{y} L = 2\pi (10)(4\pi) = 80\pi^2 \approx 789.57\text{ cm}^2 A = 2 π y ˉ L = 2 π ( 10 ) ( 4 π ) = 80 π 2 ≈ 789.57 cm 2
5. Summary Table of Trigonometry & Mensuration Formulas
Area Core Mathematical Relationship PRC ECE Board Context
Phasors V = V m cos ϕ + j V m sin ϕ \mathbf{V} = V_m \cos\phi + j V_m \sin\phi V = V m cos ϕ + j V m sin ϕ AC steady-state circuit analysis, impedance triangles
Law of Cosines c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab\cos C c 2 = a 2 + b 2 − 2 ab cos C Oblique vector addition of non-perpendicular AC sources
Napier's Rule 1 sin ( mid ) = tan ( adj 1 ) tan ( adj 2 ) \sin(\text{mid}) = \tan(\text{adj}_1) \tan(\text{adj}_2) sin ( mid ) = tan ( adj 1 ) tan ( adj 2 ) Right spherical triangle resolution for satellite links
Napier's Rule 2 sin ( mid ) = cos ( opp 1 ) cos ( opp 2 ) \sin(\text{mid}) = \cos(\text{opp}_1) \cos(\text{opp}_2) sin ( mid ) = cos ( opp 1 ) cos ( opp 2 ) Calculating great-circle arc distances and bearings
Pappus Volume V = 2 π y ˉ A plane V = 2\pi \bar{y} A_{\text{plane}} V = 2 π y ˉ A plane Toroidal transformer cores, circular waveguide bends