2.2 Trigonometry & Solid Mensuration

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: $\sin^2 \theta + \cos^2 \theta = 1 \quad | \quad 1 + \tan^2 \theta = \sec^2 \theta \quad | \quad 1 + \cot^2 \theta = \csc^2 \theta$
• Angle Sum & Difference Identities: $\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta$ $\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$
• Double-Angle Identities: $\sin(2\theta) = 2\sin\theta\cos\theta$ $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$

Euler's Identity & Phasor Representation

In AC electrical engineering, sinusoidal voltages and currents are modeled as rotating complex phasors using Euler's formula: $e^{j\theta} = \cos\theta + j\sin\theta$ An AC voltage waveform $v(t) = V_m \cos(\omega t + \phi)$ is represented in polar phasor form as $\mathbf{V} = V_m \angle \phi = V_m (\cos\phi + j\sin\phi)$.

---

2. Law of Sines & Law of Cosines

For any oblique plane triangle with sides $a, b, c$ and opposite angles $A, B, C$:

• Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
• Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$

Worked ECE Board Example 1: AC Voltage Vector Addition

Problem: Two AC voltage sources connected in series produce voltages $v_1(t) = 80 \cos(\omega t)\text{ V}$ and $v_2(t) = 60 \cos(\omega t + 60^\circ)\text{ V}$. Find the magnitude of the total peak voltage $V_T$ and its phase angle $\theta_T$.

Solution:

• Step 1: Represent as vectors $\mathbf{V}_1 = 80 \angle 0^\circ$ and $\mathbf{V}_2 = 60 \angle 60^\circ$.
• Step 2: In the vector addition triangle, the angle between $\mathbf{V}_1$ and $\mathbf{V}_2$ is $180^\circ - 60^\circ = 120^\circ$. Apply the Law of Cosines to find resultant magnitude $V_T$: $V_T^2 = 80^2 + 60^2 - 2(80)(60)\cos(120^\circ)$ $V_T^2 = 6400 + 3600 - 9600(-0.5) = 10000 + 4800 = 14800$ $V_T = \sqrt{14800} = 121.66\text{ V}$
• Step 3: Apply the Law of Sines to find phase angle $\theta_T$: $\frac{60}{\sin\theta_T} = \frac{121.66}{\sin(120^\circ)} \implies \sin\theta_T = \frac{60 \times 0.866025}{121.66} = \frac{51.9615}{121.66} = 0.4271$ $\theta_T = \arcsin(0.4271) = 25.28^\circ$ Total voltage vector: $\mathbf{V}_T = 121.66 \angle 25.28^\circ\text{ V}$.

---

3. Spherical Trigonometric Relations & Napier's Rules

Spherical trigonometry deals with triangles drawn on the surface of a sphere, where the sides $a, b, c$ 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 $C = 90^\circ$), eliminate right angle $C$ and arrange the remaining 5 parts in circular order: $a$, $b$, $\text{co-}A$ ($90^\circ - A$), $\text{co-}c$ ($90^\circ - c$), and $\text{co-}B$ ($90^\circ - B$).

Napier's Two Rules

1. Rule 1 (Sine-TaN Rule): The sine of any middle part is equal to the product of the tangents of the two adjacent parts: $\sin(\text{middle}) = \tan(\text{adjacent}_1) \cdot \tan(\text{adjacent}_2)$
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: $\sin(\text{middle}) = \cos(\text{opposite}_1) \cdot \cos(\text{opposite}_2)$

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 $a = 30^\circ$ and side $b = 45^\circ$ (where angle $C = 90^\circ$). Calculate the great-circle arc length $c$ in degrees and in kilometers (taking Earth's radius $R = 6,371\text{ km}$).

Solution:

• Step 1: Select middle part $\text{co-}c = 90^\circ - c$. Its opposite parts are $a$ and $b$.
• Step 2: Apply Napier's Rule 2 (Sine-CoS Rule): $\sin(\text{co-}c) = \cos(a) \cdot \cos(b) \implies \sin(90^\circ - c) = \cos a \cos b$ $\cos c = \cos(30^\circ) \cdot \cos(45^\circ) = (0.866025) \cdot (0.707107) = 0.612372$ $c = \arccos(0.612372) = 52.24^\circ$
• Step 3: Convert angular distance $c$ to linear distance along Earth's surface: $c_{\text{rad}} = 52.239^\circ \times \frac{\pi}{180^\circ} = 0.91174\text{ rad}$ $D = R \cdot c_{\text{rad}} = 6371\text{ km} \times 0.91174 = 5,808.7\text{ km}$

---

4. Solid Mensuration & Theorems of Pappus-Guldinus

Standard Solid Figures

• Right Circular Cone: $V = \frac{1}{3}\pi r^2 h$, Lateral Area $A_L = \pi r s$ (where slant height $s = \sqrt{r^2 + h^2}$).
• Frustum of a Cone: $V = \frac{1}{3}\pi h (R^2 + r^2 + R r)$.
• Sphere: $V = \frac{4}{3}\pi r^3$, Surface Area $A = 4\pi r^2$.

Theorems of Pappus-Guldinus for Solids of Revolution

1. 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 $L$ and the distance traveled by its centroid $\bar{y}$: $A = 2\pi \bar{y} L$
2. Second Theorem (Volume): The volume of the solid generated by revolving a plane area $A_{\text{plane}}$ about a non-intersecting axis in its plane equals the product of the area and the distance traveled by its centroid $\bar{y}$: $V = 2\pi \bar{y} A_{\text{plane}}$

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 $r = 2\text{ cm}$ around an axis located at a distance $R = 10\text{ cm}$ 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: $A_{\text{plane}} = \pi r^2 = \pi (2)^2 = 4\pi\text{ cm}^2$ $L = 2\pi r = 2\pi (2) = 4\pi\text{ cm}$
• Step 2: Centroid distance to axis of rotation is $\bar{y} = R = 10\text{ cm}$.
• Step 3: Apply Pappus's Second Theorem for Volume: $V = 2\pi \bar{y} A_{\text{plane}} = 2\pi (10)(4\pi) = 80\pi^2 \approx 789.57\text{ cm}^3$
• Step 4: Apply Pappus's First Theorem for Surface Area: $A = 2\pi \bar{y} L = 2\pi (10)(4\pi) = 80\pi^2 \approx 789.57\text{ cm}^2$

---

5. Summary Table of Trigonometry & Mensuration Formulas