Applications of Trigonometry and Vectors

Surveyors use a method known as triangulation to measure distances when direct measurements cannot be made due to obstructions in the line of sight. The topics covered in this chapter include the application of trigonometric laws in oblique triangles and vectors in various scenarios.

7.1 Oblique Triangles and the Law of Sines

Congruency and Oblique Triangles
  • Focuses on solving triangles that are not right triangles by developing laws that exist between the sides and angles of any triangle. Congruence axioms (SAS, ASA, SSS) are essential for this process.

Congruence Axioms:

  1. Side-Angle-Side (SAS): If two sides and the included angle of one triangle are equal to those of another triangle, then the triangles are congruent.
  2. Angle-Side-Angle (ASA): If two angles and the included side of one triangle are equal to those of another triangle, then the triangles are congruent.
  3. Side-Side-Side (SSS): If the three sides of one triangle are equal to the three sides of another triangle, then the triangles are congruent.
Data Required for Solving Oblique Triangles

There are four possible cases regarding known elements of the triangles:

  1. Case 1 (SAA or ASA): One side and two angles are known.
  2. Case 2 (SSA): Two sides and an angle not included are known; this may lead to zero, one, or two triangles.
  3. Case 3 (SAS): Two sides and the included angle are known.
  4. Case 4 (SSS): Three sides are known.

Note: If all three angles are given (AAA), we conclude only similarity, not congruence, which does not yield unique side lengths.

Law of Sines

The Law of Sines states that in any triangle ABC, with sides a, b, and c,
asin⁡A=bsin⁡B=csin⁡C.\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.
This indicates that the lengths of the sides of a triangle are proportional to the sines of the angles opposite them.

Derivation of the Law of Sines
  1. Construct a perpendicular from vertex B to side AC, label it h. Express side lengths in terms of h using trigonometric ratios.
  2. Set up the equations relating the sides and angles as follows:
    • h=asin⁡Ch = a \sin C
    • h=csin⁡Ah = c \sin A
    • Set two equations equal to each other:
    • asin⁡Csin⁡A=c\frac{a \sin C}{\sin A} = c
    • Repeat for the other sides to finalize the derivation of the Law of Sines.
Area of a Triangle

The area of a triangle can also be computed using the formula:
A=12bhA = \frac{1}{2} b h
Where A represents area, b is the base, and h is the height. Using the Law of Sines, alternate area calculations yield:
A=12absin⁡CA = \frac{1}{2}ab \sin C .

Solutions of SAA and ASA Triangles

Example 1: Given triangle ABC with $A = 32.0^ ext{°}$, $B = 81.8^ ext{°}$, and $a = 42.9 ext{ cm}$, apply the Law of Sines:

  1. Solve for side b using the known angles and the Law of Sines:
    asin⁡A=bsin⁡B⟹b=asin⁡Bsin⁡A\frac{a}{\sin A} = \frac{b}{\sin B} \Longrightarrow b = \frac{a \sin B}{\sin A}
  2. Find angle C using the angle sum formula:
    C=180ext°−A−BC = 180^ ext{°} - A - B
  3. Finally, use the Law of Sines again to compute side c.

Example 2: Distance across a river calculated using various triangles and bearings showcased using a similar procedure.

7.2 The Ambiguous Case of the Law of Sines

Description of the Ambiguous Case
  1. This case arises when the lengths of two sides and the angle opposite one of them (SSA) are given: there may be zero, one, or two possible triangles.
  2. Key outcomes:
    • If the known angle is acute and $a < h < b$: zero solutions.
    • If $a = h < b$: one solution (right triangle).
    • If $h < a < b$: two solutions exist.
    • If the known angle is obtuse, further constraints apply.
Solutions of SSA Triangles

Example 1: Consider triangle ABC where given $B = 55^ ext{°}$, $b = 8.94$, and $a = 25.1$. Find angle A:
sin⁡A=absin⁡B\sin A = \frac{a}{b} \sin B and evaluate for possible triangle configurations.

7.3 The Law of Cosines

Derivation of Law of Cosines
  • Applicable to both SAS and SSS configurations for uniquely determining the sides or angles of triangles through formulas:
    a2=b2+c2−2bccosAa^2 = b^2 + c^2 - 2bc cos A
  • Example: Apply Law of Cosines to compute missing triangle side lengths.
Heron's Formula for Area

A=s(s−a)(s−b)(s−c), where s=12(a+b+c)A = \sqrt{s(s-a)(s-b)(s-c)}, \text{ where } s = \frac{1}{2}(a+b+c) gives the area given three side lengths.

7.4 Geometrically Defined Vectors and Applications

Basic Terminology
  1. Vectors represent quantities with both magnitude and direction; scalar quantities have only magnitude.
  2. The vector sum can be computed geometrically and is strength in its resulting direction.
  3. The concept of equilibriums introduces how opposing vectors can balance out forces.
Applications with Force Problems
  • Analyzing vectors to determine forces acting on objects moving on slopes or through fluid mediums, using trigonometric foundations for stability through mathematical relationships.
  • Example: Finding resultant force magnitudes using vector decompositions, or applying the law of cosines/sines for angle calculations in force diagrams involved.

7.5 Algebraically Defined Vectors and the Dot Product

Operations and Definitions
  • Discuss properties of vectors and computations that determine angles, scalar products, and vector equivalences.
  • Example: Finding angles between vectors through dot product relationships, yielding insights regarding their respective orientations and physical alignments.
Relationships in Vectors
  • Equating angles provides insights into geometric configurations in fields concerned with physics, navigation, engineering, etc.