6.3 Practical Measurement, Optimization, and Trigonometric Modeling

6.3 Practical Measurement, Optimization, and Trigonometric Modeling

Real-world applications and optimization problems (最值与范围问题) represent the highest tier of difficulty in the Gaokao triangle module. These problems test a student's ability to extract geometric models from textual scenarios, set up single-variable trigonometric functions, and apply algebraic inequalities under strict geometric domain constraints.

---

1. Terminology and Models in Practical Measurement

Directional and Spatial Angle Definitions

1. Angle of Elevation (仰角): The angle between the line of sight and the horizontal plane when looking upward at a target.
2. Angle of Depression (俯角): The angle between the line of sight and the horizontal plane when looking downward at a target.
3. Bearing Angle (方位角): The angle measured clockwise starting from due North ($0^\circ$) to the line of direction, ranging from $0^\circ$ to $360^\circ$.
4. Direction Angle (方向角): Direction specified relative to cardinal directions, e.g., "North $30^\circ$ East" (北偏东 $30^\circ$), meaning rotated $30^\circ$ toward East from due North.

                  North (0° / 360°)
                        |
                        |   / Target (North 30° East)
                        |  / 
                        | / 30°
         West ----------+---------- East (90°)
                        |
                        |
                  South (180°)

Standard 2D and 3D Surveying Structures

• Inaccessible Distance Model: To measure distance between target points $A$ and $B$ across a river, select a baseline $CD$ on one bank, measure length $CD = d$, and observe angles $\angle ACD, \angle BCD, \angle ADC, \angle BDC$. Solve $\triangle ACD$ and $\triangle BCD$ to get $AC$ and $BC$, then apply the Cosine Theorem in $\triangle ABC$.
• Height Measurement Model: To measure peak height $h = TT'$ above horizontal ground, set two observation points $A$ and $B$ aligned with baseline $T'$. Measure distance $AB = d$, elevation angles $\alpha = \angle TAT'$ and $\beta = \angle TBT'$. Apply the Sine Theorem in vertical triangle $\triangle TAB$.

---

2. Optimization Techniques for Triangle Perimeter and Area

When a side $a$ and its opposite angle $A$ are fixed in $\triangle ABC$, the triangle is partially constrained, causing perimeter $P = a + b + c$ and area $S = \frac{1}{2}bc \sin A$ to vary. Two main analytical methods find their extrema:

Method I: Algebraic Inequality Method (AM-GM Inequality)

Applying the Cosine Theorem to fixed $a$ and $A$: $a^2 = b^2 + c^2 - 2bc \cos A$ Since $b^2 + c^2 \ge 2bc$ by AM-GM: $a^2 \ge 2bc - 2bc \cos A = 2bc (1 - \cos A)$ Rearranging gives an explicit upper bound for product $bc$: $bc \le \frac{a^2}{2(1 - \cos A)}$ Equality holds if and only if $b = c$. Thus, maximum area $S_{max}$ is: $S_{max} = \frac{1}{2} bc_{max} \sin A = \frac{a^2 \sin A}{4(1 - \cos A)} = \frac{a^2}{4 \tan(A/2)}$

Method II: Trigonometric Single-Variable Substitution

By the Sine Theorem, $b = 2R \sin B$ and $c = 2R \sin C$, where $2R = \frac{a}{\sin A}$. Express $b + c$ as a function of angle $B$: $b + c = 2R (\sin B + \sin C) = 2R \left( \sin B + \sin(\pi - A - B) \right)$ Using sum-to-product or expanding $\sin(\pi - A - B) = \sin(A + B)$: $b + c = 2R \left( \sin B + \sin A \cos B + \cos A \sin B \right) = 2R \left( (1 + \cos A)\sin B + \sin A \cos B \right)$ By auxiliary angle transformation $C_0 \sin(B + \phi)$: $b + c = 4R \sin\left(\frac{B+C}{2}\right) \cos\left(\frac{B-C}{2}\right) = 4R \cos\left(\frac{A}{2}\right) \cos\left(\frac{B-C}{2}\right)$ Since $\cos\left(\frac{B-C}{2}\right) \le 1$ with equality when $B = C$, the maximum sum of sides is $4R \cos\left(\frac{A}{2}\right)$.

---

3. Worked Gaokao Exam Examples

Example 1 (Perimeter Range and Extremum Optimization)

In $\triangle ABC$, angle $A = \frac{\pi}{3}$ ($60^\circ$) and opposite side $a = \sqrt{3}$.

1. Determine the circumradius $R$.
2. Find the range of the perimeter $P = a + b + c$ and state its maximum value.

Solution:

Part 1: By the Sine Theorem: $2R = \frac{a}{\sin A} = \frac{\sqrt{3}}{\sin\frac{\pi}{3}} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2 \implies R = 1$

Part 2: Express $b + c$ in terms of angle $B$: $b = 2R \sin B = 2 \sin B$ $c = 2R \sin C = 2 \sin\left(\pi - A - B\right) = 2 \sin\left(\frac{2\pi}{3} - B\right)$ Expand $c$: $c = 2 \left( \sin\frac{2\pi}{3} \cos B - \cos\frac{2\pi}{3} \sin B \right) = 2 \left( \frac{\sqrt{3}}{2} \cos B - \left(-\frac{1}{2}\right) \sin B \right) = \sqrt{3} \cos B + \sin B$ Summing $b + c$: $b + c = 2 \sin B + \sin B + \sqrt{3} \cos B = 3 \sin B + \sqrt{3} \cos B$ Apply the auxiliary angle formula $A_0 \sin(B + \phi)$ where $A_0 = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{12} = 2\sqrt{3}$: $b + c = 2\sqrt{3} \left( \frac{\sqrt{3}}{2} \sin B + \frac{1}{2} \cos B \right) = 2\sqrt{3} \sin\left(B + \frac{\pi}{6}\right)$ Determine the domain of $B$: since $A = \frac{\pi}{3}$ and $A + B + C = \pi$, we have $0 < B < \frac{2\pi}{3}$. Adding $\frac{\pi}{6}$ to all terms: $\frac{\pi}{6} < B + \frac{\pi}{6} < \frac{5\pi}{6}$ Within this interval, the sine function satisfies $\frac{1}{2} < \sin\left(B + \frac{\pi}{6}\right) \le 1$. Multiplying by $2\sqrt{3}$: $\sqrt{3} < b + c \le 2\sqrt{3}$ Since $a = \sqrt{3}$, the perimeter $P = a + b + c = \sqrt{3} + (b + c)$ has domain: $2\sqrt{3} < P \le 3\sqrt{3}$ Thus, the range of perimeter $P$ is $(2\sqrt{3}, 3\sqrt{3}]$, and its maximum value is $3\sqrt{3}$ attained when $B + \frac{\pi}{6} = \frac{\pi}{2} \implies B = \frac{\pi}{3} = C$ (equilateral triangle).

---

Example 2 (Real-World Engineering Surveying Model)

Surveyors at point $A$ observe a target tower $T$ on top of a mountain. Point $A$ is on horizontal ground. The angle of elevation of point $A$ to the mountain base $B$ is $30^\circ$, and to the top of tower $T$ is $45^\circ$. The surveyors walk $100$ meters along a straight slope of incline $30^\circ$ directly toward base $B$ to reach point $C$. From point $C$, the angle of elevation to tower top $T$ is measured to be $60^\circ$. Find the height of the tower $T$ if $B$ and $T$ are vertically aligned.

Solution:

Let the horizontal plane pass through $A$. Path $AC = 100$ m lies along line $AB$ inclined at $30^\circ$. In vertical triangle $\triangle ATC$:

1. $\angle TAC = 45^\circ - 30^\circ = 15^\circ$.
2. $\angle TCA = 180^\circ - (60^\circ - 30^\circ) = 150^\circ$.
3. $\angle ATC = 180^\circ - 15^\circ - 150^\circ = 15^\circ$. Since $\angle TAC = \angle ATC = 15^\circ$, $\triangle ATC$ is an isosceles triangle with $TC = AC = 100$ meters! Now consider right triangle formed by top $T$, base $B$, and point $C$. Using $\triangle TBC$, height $TB$ is found directly: $TB = TC \cdot \sin 60^\circ = 100 \cdot \frac{\sqrt{3}}{2} = 50\sqrt{3} \text{ meters}$ The tower/mountain structure height is $50\sqrt{3}$ meters.