6.2 Right Triangle Trigonometry and Geometry in Surveying

6.2 Right Triangle Trigonometry and Geometry in Surveying

Plane surveying assumes that measurements occur on a flat horizontal plane where the laws of Euclidean geometry apply. Right-triangle trigonometry and elementary geometry form the core computational engine for field technicians. Whether reducing electronic distance measurement (EDM) slope distances to horizontal components, determining trigonometric elevations, or verifying the angular closure of a multi-sided traverse, a solid grasp of geometric relationships is mandatory.

---

Right Triangle Trigonometry (SOH CAH TOA)

A right triangle contains one $90^\circ$ angle. The side opposite the $90^\circ$ right angle is the hypotenuse ($c$), which is always the longest side. Relative to an acute reference angle $\theta$:

• Opposite side ($a$): The side facing directly across from angle $\theta$.
• Adjacent side ($b$): The side adjacent to angle $\theta$ (that is not the hypotenuse).

The Three Primary Trigonometric Ratios

The mnemonic SOH CAH TOA defines the basic trigonometric functions:

1. Sine (SOH): $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{a}{c} \implies a = c \cdot \sin(\theta)$
2. Cosine (CAH): $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{b}{c} \implies b = c \cdot \cos(\theta)$
3. Tangent (TOA): $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{a}{b} \implies a = b \cdot \tan(\theta)$

Finding Unknown Angles Using Inverse Trigonometric Functions

When two side lengths are known, inverse trigonometric functions (arcsin, arccos, arctan) yield the unknown angle $\theta$: $\theta = \arcsin\left(\frac{\text{Opposite}}{\text{Hypotenuse}}\right) = \arccos\left(\frac{\text{Adjacent}}{\text{Hypotenuse}}\right) = \arctan\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)$

---

The Pythagorean Theorem ($a^2 + b^2 = c^2$)

In any right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: $a^2 + b^2 = c^2$

Solving for individual sides: $c = \sqrt{a^2 + b^2}, \quad a = \sqrt{c^2 - b^2}, \quad b = \sqrt{c^2 - a^2}$

Slope Distance Reduction in Surveying

When a total station measures a slope distance ($SD$) between the instrument center and a prism, and the difference in elevation ($\Delta VD$) between the two points is known from differential leveling or vertical sensor readings, the horizontal distance ($HD$) is calculated directly using the Pythagorean theorem:

$HD = \sqrt{SD^2 - \Delta VD^2}$

Step-by-Step Example: Pythagorean Reduction

Problem: A total station measures a slope distance ($SD$) of 485.22 feet to a prism benchmark. The vertical elevation difference ($\Delta VD$) between the instrument center and the prism is 32.15 feet. Calculate the horizontal distance ($HD$).

1. Identify the triangle sides:
   • Hypotenuse $c = SD = 485.22 \text{ ft}$
   • Vertical leg $a = \Delta VD = 32.15 \text{ ft}$
   • Horizontal leg $b = HD = ?$
2. Apply the Pythagorean Theorem: $HD = \sqrt{485.22^2 - 32.15^2}$
3. Calculate squares: $485.22^2 = 235,438.4484$ $32.15^2 = 1,033.6225$
4. Subtract and take square root: $HD = \sqrt{235,438.4484 - 1,033.6225} = \sqrt{234,404.8259} = 484.1537 \text{ feet}$
5. Round appropriately: The horizontal distance is 484.15 feet.

---

Geometry of Polygons: Interior Angles Formula

A closed traverse forms a polygon with $n$ sides (and $n$ interior angles). The theoretical sum of all interior angles ($S$) in any non-self-intersecting closed polygon is determined by the formula:

$S = (n - 2) \times 180^\circ$

Where $n$ represents the number of sides or vertices of the polygon.

Interior Angle Sums for Common Polygons

Traverse Angular Misclosure: When a field crew measures all interior angles of a closed boundary traverse, the sum of the measured angles will rarely equal the exact theoretical sum due to small observational errors. The difference between the measured sum and $(n - 2) \times 180^\circ$ is the angular misclosure, which must be checked against allowable field tolerances prior to adjusting the traverse.

---

Angle Relationships: Complementary, Supplementary & Zenith Angles

Field geometric proofs and instrument reductions rely on key angle definitions:

1. Complementary Angles: Two angles whose sum equals $90^\circ$ ($\theta_1 + \theta_2 = 90^\circ$).
   • In a right triangle, the two acute angles are always complementary.
2. Supplementary Angles: Two angles whose sum equals $180^\circ$ ($\theta_1 + \theta_2 = 180^\circ$).
   • Interior and exterior deflection angles along a straight baseline are supplementary.
3. Opposite (Vertical) Angles: When two straight lines intersect, the non-adjacent angles facing each other are equal (congruent).
4. Zenith Angle vs. Vertical Angle:
   • Zenith Angle ($Z$): Measured from $0^\circ$ directly overhead (zenith).
   • Vertical Angle ($V$): Measured from $0^\circ$ on the horizontal plane (positive above horizontal, negative below).
   • Relationship: $V = 90^\circ - Z$ or $Z + V = 90^\circ$.
   • Zenith angle and vertical angle are complementary!

$\text{Horizontal Distance } HD = SD \cdot \sin(Z) = SD \cdot \cos(V)$ $\text{Vertical Distance } \Delta VD = SD \cdot \cos(Z) = SD \cdot \sin(V)$

---

Practical Surveying Problem: Trigonometric Height Determination

Problem: A survey crew must determine the height of an inaccessible power line structure above a benchmark point. The total station is set up at Point A with an instrument height ($HI$) of 5.30 feet. The horizontal distance from Point A to the base of the structure is measured as 215.40 feet. The observed vertical angle to the top of the structure is $+18^\circ 24' 30''$. The rod reading at the base of the structure is 5.30 feet (equal to HI). Calculate the height of the power line structure.

1. Convert Vertical Angle to Decimal Degrees: $V = 18 + \frac{24}{60} + \frac{30}{3600} = 18 + 0.4000 + 0.008333 = 18.408333^\circ$
2. Apply Tangent Ratio (TOA): $\tan(V) = \frac{\Delta VD}{HD} \implies \Delta VD = HD \cdot \tan(V)$
3. Compute Vertical Height Difference ($\Delta VD$): $\Delta VD = 215.40 \text{ ft} \times \tan(18.408333^\circ) = 215.40 \times 0.332827 = 71.691 \text{ feet}$
4. Determine Building Height Above Base: Since the instrument $HI$ (5.30 ft) equals the target reading at the ground base (5.30 ft), the vertical elevation difference $\Delta VD$ represents the net height above ground. The total height of the structure is 71.69 feet.