Comprehensive Study Guide: Right Triangle Trigonometry, Applications, and Geometric Vectors

Right Triangle Trigonometry

• Definition II (Right Triangle Ratios):
  • Let triangle $ABC$ be a right triangle with right angle $C = 90^\circ$.
  • Angle $A$ has opposite side $a$, adjacent side $b$, and hypotenuse $c$.
  • Side $a$ is also adjacent to angle $B$, and side $b$ is also opposite to angle $B$.

• The six trigonometric functions of acute angle $A$ are defined as:

  • Sine: $\sin(A) = \frac{\text{side opposite } A}{\text{hypotenuse}} = \frac{a}{c}$
  • Cosine: $\cos(A) = \frac{\text{side adjacent } A}{\text{hypotenuse}} = \frac{b}{c}$
  • Tangent: $\tan(A) = \frac{\text{side opposite } A}{\text{side adjacent } A} = \frac{a}{b}$
  • Cotangent: $\cot(A) = \frac{\text{side adjacent } A}{\text{side opposite } A} = \frac{b}{a}$
  • Secant: $\sec(A) = \frac{\text{hypotenuse}}{\text{side adjacent } A} = \frac{c}{b}$
  • Cosecant: $\csc(A) = \frac{\text{hypotenuse}}{\text{side opposite } A} = \frac{c}{a}$

• Example: Finding Trigonometric Ratios from Given Sides

  • Given right triangle $ABC$ with $C = 90^\circ$, $a = 6$, and $c = 10$.
  • Step 1: Calculate the missing side $b$ using the Pythagorean Theorem:     $a^2 + b^2 = c^2$$6^2 + b^2 = 10^2$$36 + b^2 = 100$$b^2 = 64$$b = 8$
  • Step 2: Calculate all six trigonometric ratios for angle $A$:     $\sin(A) = \frac{6}{10} = \frac{3}{5}$$\cos(A) = \frac{8}{10} = \frac{4}{5}$$\tan(A) = \frac{6}{8} = \frac{3}{4}$$\cot(A) = \frac{8}{6} = \frac{4}{3}$$\sec(A) = \frac{10}{8} = \frac{5}{4}$$\csc(A) = \frac{10}{6} = \frac{5}{3}$

• Trigonometric Ratios for Angle B and Cofunction Relations:

  • Applying Definition II to angle $B$ yields:     $\sin(B) = \frac{\text{side opposite } B}{\text{hypotenuse}} = \frac{b}{c} = \cos(A)$$\cos(B) = \frac{\text{side adjacent } B}{\text{hypotenuse}} = \frac{a}{c} = \sin(A)$$\tan(B) = \frac{\text{side opposite } B}{\text{side adjacent } B} = \frac{b}{a} = \cot(A)$$\cot(B) = \frac{\text{side adjacent } B}{\text{side opposite } B} = \frac{a}{b} = \tan(A)$$\sec(B) = \frac{\text{hypotenuse}}{\text{side adjacent } B} = \frac{c}{a} = \csc(A)$$\csc(B) = \frac{\text{hypotenuse}}{\text{side opposite } B} = \frac{c}{b} = \sec(A)$

• Cofunction Pairs:

  • Sine and cosine are cofunctions.
  • Tangent and cotangent are cofunctions.
  • Secant and cosecant are cofunctions.

• Cofunction Theorem:

  • A trigonometric function of an angle is always equal to the cofunction of the complement of the angle.
  • Complementary angles sum to $90^\circ$ ($A + B = 90^\circ$).

• Mathematical statement of cofunction identities:     $\text{If } A + B = 90^\circ, \text{ then } \begin{cases} \sin(A) = \cos(B) \\ \sec(A) = \csc(B) \\ \tan(A) = \cot(B) \end{cases}$

• Specific example with $40^\circ$ and $50^\circ$:     $\sin(40^\circ) = \cos(50^\circ)$$\sec(40^\circ) = \csc(50^\circ)$$\tan(40^\circ) = \cot(50^\circ)$

  • Cofunction Fill-in-the-Blank Problems:

• Statement 1: $\sin(\underline{60^\circ}) = \cos(30^\circ)$, since $90^\circ - 30^\circ = 60^\circ$.

• Statement 2: $\tan(y) = \cot(\underline{90^\circ - y})$, since the complement of $y$ is $90^\circ - y$.

• Statement 3: $\sec(75^\circ) = \csc(\underline{15^\circ})$, since $90^\circ - 75^\circ = 15^\circ$.

  • Special Right Triangles:

• $30^\circ-60^\circ-90^\circ$ Triangle (with shortest side $t = 1$):

  • Side opposite $30^\circ$ angle: $1$
  • Side opposite $60^\circ$ angle: $\sqrt{3}$
  • Hypotenuse: $2$

• $45^\circ-45^\circ-90^\circ$ Triangle (with legs $t = 1$):

  • Side opposite $45^\circ$ angles: $1$

  • Hypotenuse: $\sqrt{2}$

  • Exact Values of Special Angles:

• Values for $30^\circ$:     $\sin(30^\circ) = \frac{1}{2}$$\cos(30^\circ) = \frac{\sqrt{3}}{2}$$\tan(30^\circ) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

• Values for $45^\circ$:     $\sin(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$\cos(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$\tan(45^\circ) = 1$

• Values for $60^\circ$:     $\sin(60^\circ) = \frac{\sqrt{3}}{2}$$\cos(60^\circ) = \frac{1}{2}$$\tan(60^\circ) = \sqrt{3}$

  • Complete Exact Value Summary Table ($0^\circ$ to $90^\circ$):

• Pattern for Sine: $\sin(\theta) = \frac{\sqrt{n}}{2}$ for $n = 0, 1, 2, 3, 4$.     $\sin(0^\circ) = \frac{\sqrt{0}}{2} = 0$$\sin(30^\circ) = \frac{\sqrt{1}}{2} = \frac{1}{2}$$\sin(45^\circ) = \frac{\sqrt{2}}{2}$$\sin(60^\circ) = \frac{\sqrt{3}}{2}$$\sin(90^\circ) = \frac{\sqrt{4}}{2} = 1$

• Pattern for Cosine: $\cos(\theta) = \frac{\sqrt{n}}{2}$ for $n = 4, 3, 2, 1, 0$.     $\cos(0^\circ) = \frac{\sqrt{4}}{2} = 1$$\cos(30^\circ) = \frac{\sqrt{3}}{2}$$\cos(45^\circ) = \frac{\sqrt{2}}{2}$$\cos(60^\circ) = \frac{\sqrt{1}}{2} = \frac{1}{2}$$\cos(90^\circ) = \frac{\sqrt{0}}{2} = 0$

• Values for Tangent:     $\tan(0^\circ) = 0$$\tan(30^\circ) = \frac{\sqrt{3}}{3}$$\tan(45^\circ) = 1$$\tan(60^\circ) = \sqrt{3}$$\tan(90^\circ) = \text{undefined}$

  • Example: Simplifying Expressions with Special Angle Values

• Given $x = 30^\circ$ and $y = 45^\circ$:

• Part a: Evaluate $2\sin(x)$$2\sin(30^\circ) = 2 \times \frac{1}{2} = 1$

• Part b: Evaluate $\sin(2y)$$\sin(2 \times 45^\circ) = \sin(90^\circ) = 1$

• Part c: Evaluate $4\sin(3x - 90^\circ)$$4\sin(3(30^\circ) - 90^\circ) = 4\sin(90^\circ - 90^\circ) = 4\sin(0^\circ) = 4 \times 0 = 0$

Calculators and Trigonometric Functions of Acute Angles

• Subdivisions of a Degree:

  • One degree ($1^\circ$) is defined as $\frac{1}{360}$ of a complete revolution.
  • One minute ($1'$) is $\frac{1}{60}$ of a degree:     $1^\circ = 60' \quad \text{or} \quad 1' = \left(\frac{1}{60}\right)^\circ$
  • One second ($1''$) is $\frac{1}{60}$ of a minute:     $1' = 60'' \quad \text{or} \quad 1'' = \left(\frac{1}{60}\right)'$

• Example: Addition of Angle Measures in Degrees and Minutes

  • Problem: Add $48^\circ 49'$ and $72^\circ 26'$.
  • Step 1: Add the minutes:     $49' + 26' = 75'$
  • Step 2: Convert $75'$ to degrees and minutes:     $75' = 60' + 15' = 1^\circ 15'$
  • Step 3: Add the degrees including the carried degree:     $48^\circ + 72^\circ + 1^\circ = 121^\circ$
  • Final Result: $121^\circ 15'$

• Decimal Degrees Conversion:

  • Decimal degrees express fractional parts of a degree as decimals (e.g., $30.5^\circ$, $101.75^\circ$, $62.831^\circ$).
  • Conversion procedure from decimal degrees to degrees and minutes: Multiply the fractional decimal part by $60$ to determine minutes.
  • Example: Convert $27.25^\circ$ to degrees and minutes:
  • Whole degree part: $27^\circ$
  • Fractional part to minutes: $0.25 \times 60' = 15'$
  • Final Result: $27^\circ 15'$

• Decimal Degree to Minutes Conversion Reference:

• $0.1^\circ = 6'$

• $0.2^\circ = 12'$

• $0.3^\circ = 18'$

• $0.4^\circ = 24'$

• $0.5^\circ = 30'$

• $0.6^\circ = 36'$

• $0.7^\circ = 42'$

• $0.8^\circ = 48'$

• $0.9^\circ = 54'$

• $1.0^\circ = 60'$

  • Calculator Evaluation Example:

• Problem: Calculate $\tan(58.75^\circ)$.

• Scientific Calculator keystroke sequence: 58.75 followed by tan key.

• Graphing Calculator keystroke sequence: tan ( 58.75 ) ENTER.

• Rounded to four decimal places:     $\tan(58.75^\circ) = 1.6479$

Solving Right Triangles

• Definition of Significant Digits (Figures):

  • Significant digits are counted from left to right beginning with the first non-zero digit on the left.
  • When no decimal point is explicitly present, trailing zeros are not significant.
  • Counting Examples:
  • $0.042$ has two significant digits.
  • $0.005$ has one significant digit.
  • $20.5$ has three significant digits.
  • $6.000$ has four significant digits.
  • $9,200.$ has four significant digits (decimal point present).
  • $700$ has one significant digit (no decimal point).

• Accuracy Standard Between Sides and Angles:

• Side accuracy of 2 significant digits corresponds to angle accuracy to the nearest degree.

• Side accuracy of 3 significant digits corresponds to angle accuracy to the nearest 10 minutes ($10'$) or tenth of a degree ($0.1^\circ$).

• Side accuracy of 4 significant digits corresponds to angle accuracy to the nearest minute ($1'$) or hundredth of a degree ($0.01^\circ$).

  • Right Triangle Conventions and Rounding Rules:

• Angle $C$ is designated as the $90^\circ$ right angle unless stated otherwise.

• Final calculated values are rounded to match the number of significant digits in the least significant number provided in the initial problem statement.

• Intermediate step calculations must maintain unrounded precision to avoid compounding rounding errors.

  • Example: Solving a Right Triangle Given Two Sides

• Problem: Right triangle $ABC$ with $C = 90^\circ$, $a = 2.73$, and $b = 3.41$. Find side $c$, angle $A$, and angle $B$.

• Step 1: Find hypotenuse $c$ using the Pythagorean Theorem:     $c = \sqrt{a^2 + b^2} = \sqrt{2.73^2 + 3.41^2} = \sqrt{7.4529 + 11.6281} = \sqrt{19.081} \approx 4.3682$     Rounding to 3 significant digits gives $c = 4.37$

• Step 2: Find angle $A$ using tangent:     $\tan(A) = \frac{a}{b} = \frac{2.73}{3.41} \approx 0.800586$$A = \arctan(0.800586) \approx 38.682^\circ$     Rounding to nearest tenth of a degree gives $A = 38.7^\circ$

• Step 3: Find angle $B$ using complementary property:     $B = 90^\circ - 38.682^\circ = 51.318^\circ$     Rounding to nearest tenth of a degree gives $B = 51.3^\circ$

  • Example: Ferris Wheel Application

• Specifications of original 1893 Ferris wheel built by George W. G. Ferris:
  • Wheel diameter = $250\,\text{ft}$ (wheel radius $r = 125\,\text{ft}$).
  • Total height of top of wheel above ground = $264\,\text{ft}$.
  • Ground clearance at bottom point $P_0$: $264\,\text{ft} - 250\,\text{ft} = 14\,\text{ft}$.
  • Height of wheel center $O$ above ground: $14\,\text{ft} + 125\,\text{ft} = 139\,\text{ft}$.
  • Wheel capacity: 36 cars holding 40 passengers each; 20 minutes per revolution.
• Geometry for rider position at central angle $\theta = 45^\circ$ from bottom $P_0$ to $P_1$:
  • Let $x$ be the vertical distance from center $O$ down to position $P_1$:       $\cos(45^\circ) = \frac{x}{125}$$x = 125 \cos(45^\circ) = 125 \times 0.707107 \approx 88.388\,\text{ft}$
  • Rider height $h$ above ground at $P_1$:       $h = (\text{height of center } O) - x$$h = 139 - 88.388 = 50.612\,\text{ft}$
  • Expressed to appropriate precision, height $h = 51\,\text{ft}$ (or $50.6\,\text{ft}$).

Practical Applications of Right Triangle Trigonometry

• Example: Geometry of an Isosceles Triangle

  • Problem: An isosceles triangle has two equal sides of length $24\,\text{cm}$ and equal base angles of $52^\circ$. Find the altitude and base length.
  • Step 1: Draw altitude $h$ from apex to base, forming two congruent right triangles with hypotenuse $24\,\text{cm}$ and acute angle $52^\circ$.
  • Step 2: Calculate altitude $h$:     $\sin(52^\circ) = \frac{h}{24}$$h = 24 \sin(52^\circ) = 24 \times 0.78801 = 18.912\,\text{cm} \approx 19\,\text{cm}$
  • Step 3: Calculate half of base length $d$:     $\cos(52^\circ) = \frac{d}{24}$$d = 24 \cos(52^\circ) = 24 \times 0.61566 = 14.776\,\text{cm}$
  • Step 4: Calculate total base length $b = 2d$:     $b = 2 \times 14.776 = 29.552\,\text{cm} \approx 30\,\text{cm}$

• Definitions of Angle of Elevation and Angle of Depression:

• Angle of Elevation: An angle measured upward from a horizontal baseline to the line of sight.

• Angle of Depression: An angle measured downward from a horizontal baseline to the line of sight.

• Both angles are non-negative real quantities defined relative to horizontal lines.

• Line of sight: The straight path connecting the observer's eye (vertex) to the observed object.

  • Example: Height and Shadow Angle of Elevation

• Problem: A $75.0\,\text{ft}$ vertical flagpole casts a shadow $43.0\,\text{ft}$ long on horizontal ground. Find the sun's angle of elevation $\theta$ to the nearest 10 minutes.

• Step 1: Set up tangent ratio:     $\tan(\theta) = \frac{\text{height}}{\text{shadow}} = \frac{75.0}{43.0} \approx 1.744186$

• Step 2: Evaluate arctangent:     $\theta = \arctan(1.744186) \approx 60.169^\circ$

• Step 3: Convert fractional degree to minutes:     $0.169 \times 60' = 10.14' \approx 10'$

• Final Angle of Elevation: $\theta = 60^\circ 10'$

  • Definition of Bearing in Navigation and Surveying:

• The bearing of a line $l$ is the acute angle formed between the north-south directional line and line $l$.

• Notation format: Begins with primary direction $\text{N}$ (North) or $\text{S}$ (South), followed by angle magnitude in degrees, and ends with secondary direction $\text{E}$ (East) or $\text{W}$ (West).

• Examples: $\text{N}\, 30^\circ\, \text{E}$, $\text{S}\, 45^\circ\, \text{W}$, $\text{N}\, 13^\circ\, \text{E}$, $\text{S}\, 19^\circ\, \text{E}$.

  • Example: Navigation Bearing Calculation

• Problem: San Luis Obispo (S.L.O.) is $12\,\text{mi}$ due north of Grover Beach (G.B.). Arroyo Grande (A.G.) is $4.6\,\text{mi}$ due east of Grover Beach. Find the bearing of San Luis Obispo from Arroyo Grande.

• Step 1: Model the right triangle formed by points S.L.O., G.B., and A.G., where angle at G.B. is $90^\circ$.

• Step 2: Angle $\theta$ at S.L.O. inside triangle equals angle measured west from North line at A.G. (alternate interior angles):     $\tan(\theta) = \frac{\text{G.B. to A.G.}}{\text{G.B. to S.L.O.}} = \frac{4.6}{12} \approx 0.38333$

• Step 3: Calculate angle $\theta$:     $\theta = \arctan(0.38333) \approx 20.97^\circ \approx 21^\circ$

• Final Bearing of San Luis Obispo from Arroyo Grande: $\text{N}\, 21^\circ\, \text{W}$

  • Example: Distance and Position Verification (Helicopter Search Scenario)

• Problem Statement:
  • Radar stations $A$ and $B$ are $25\,\text{mi}$ apart on a straight N-S road.
  • Helicopter lands at point $C$. Bearing from $A$ is $\text{N}\, 13^\circ\, \text{E}$; bearing from $B$ is $\text{S}\, 19^\circ\, \text{E}$.
  • Station instructs pilot to walk due west $3.5\,\text{mi}$ to reach the N-S road at point $F$.
  • Task: Verify if $3.5\,\text{mi}$ distance is correct.
• Mathematical Proof and Solution:
  • Let $AB = 25\,\text{mi}$ along the N-S axis, with $F$ being the perpendicular foot of $C$ on line $AB$.
  • Let distance $CF = y$ (perpendicular distance due west to road).
  • Let distance $AF = x$, making distance FB = 25 - x$.\n - In right triangle AFC:\n      \tan(13^\circ) = \frac{y}{x} \implies y = x \tan(13^\circ)\n - In right triangle BFC:\n      \tan(19^\circ) = \frac{y}{25 - x} \implies y = (25 - x) \tan(19^\circ)\n - Equating expressions for y:\n      x \tan(13^\circ) = (25 - x) \tan(19^\circ)\n      x \tan(13^\circ) + x \tan(19^\circ) = 25 \tan(19^\circ)\n      x (\tan(13^\circ) + \tan(19^\circ)) = 25 \tan(19^\circ)\n - Substituting numeric values (\tan(13^\circ) \approx 0.23087$,$\tan(19^\circ) \approx 0.34433):\n      x (0.23087 + 0.34433) = 25 (0.34433)\n      0.57520 x = 8.60825\n      x = \frac{8.60825}{0.57520} \approx 14.966\,\text{mi}\n - Calculate perpendicular distance y:\n      y = 14.966 \times \tan(13^\circ) = 14.966 \times 0.23087 \approx 3.455\,\text{mi}\n - Rounding to two significant digits yields y = 3.5\,\text{mi}.\n - Conclusion: The station's instruction for the pilot to walk 3.5\,\text{mi} due west is correct.\n\n# Geometric Vectors and Vector Components\n\n- Definition of Vector and Scalar Quantities:\n - Vector Quantity: Physical quantity defined by both magnitude and direction (e.g., velocity, acceleration, force, displacement).\n - Scalar Quantity: Physical quantity defined by magnitude only (e.g., speed, distance, time, mass, temperature).\n - Geometric Representation: Vectors are represented geometrically by directed line segments (arrows). Arrow direction indicates vector direction; arrow length indicates vector magnitude.\n\n- Vector Notation Summary:\n\n\n\n - Boldface text \mathbf{V}$or overhead arrow$\vec{\mathbf{V}} designates a vector.\n - Overhead arrow connecting points \vec{\mathbf{AB}}$designates a directed vector from initial point$A$to terminal point$B.\n - Absolute value bars around vector |\mathbf{V}|$designate the magnitude of vector$\mathbf{V}, which is a scalar quantity.\n - Italicized lowercase letter x designates a scalar variable.\n\n- Zero Vector and Vector Equivalence:\n - Zero Vector (denoted \mathbf{0}$): A vector with magnitude equal to zero ($|\mathbf{0}| = 0) and no defined direction.\n - Vector Equivalence: Two vectors are equivalent if and only if they possess identical magnitudes and identical directions, regardless of their position in coordinate space.\n\n\n\n - Opposite Vectors: Vectors with equal magnitudes but opposite directions (\mathbf{V}_4 = -\mathbf{V}_3).\n\n- Vector Addition and Subtraction:\n - Resultant Vector: The sum of vectors \mathbf{U} + \mathbf{V}.\n - Head-to-Tail Addition Method: Place the tail of vector \mathbf{V}$at the tip of vector$\mathbf{U}$. The resultant$\mathbf{U} + \mathbf{V}$extends from the tail of$\mathbf{U}$to the tip of$\mathbf{V}.\n\n\n\n - Parallelogram Addition Method: Join tails of \mathbf{U}$and$\mathbf{V} together to form adjacent sides of a parallelogram. The resultant vector is the diagonal starting from the common tail.\n - Vector Subtraction: Defined as addition of the opposite vector:\n    \mathbf{U} - \mathbf{V} = \mathbf{U} + (-\mathbf{V})\n\n\n\n - Direct Construction for Subtraction: Place tails of \mathbf{U}$and$\mathbf{V}$together; the difference vector$\mathbf{U} - \mathbf{V}$points from tip of$\mathbf{V}$to tip of$\mathbf{U}.\n\n- Example: Boat Navigation and Current Resultant Course\n - Problem: A boat heads due east at 12\,\text{mph}$relative to water across a river flowing due north at$5.1\,\text{mph}. Find the boat's actual course (speed and bearing) to two significant digits.\n - Step 1: Calculate resultant speed (magnitude |\mathbf{V}|):\n    |\mathbf{V}| = \sqrt{12^2 + 5.1^2} = \sqrt{144 + 26.01} = \sqrt{170.01} \approx 13.038\,\text{mph} \approx 13\,\text{mph}\n - Step 2: Calculate direction angle \theta north of east:\n    \tan(\theta) = \frac{5.1}{12} = 0.425\n    \theta = \arctan(0.425) \approx 23.025^\circ \approx 23^\circ\n - Step 3: Express direction as a navigational bearing:\n    \text{Bearing} = 90^\circ - 23^\circ = 67^\circ \implies \text{N}\, 67^\circ\, \text{E}\n\n- Vector Components Formulas:\n\n\n\n - Any vector \mathbf{V}$in standard position with direction angle$\theta$measured counterclockwise from the positive x-axis can be resolved into perpendicular component vectors$\mathbf{V}_x$and$\mathbf{V}_y:\n    \mathbf{V}_x = |\mathbf{V}| \cos(\theta) \quad \text{and} \quad \mathbf{V}_y = |\mathbf{V}| \sin(\theta)\n\n- Example: Human Cannonball Velocity Components\n - Problem: A human cannonball is shot with initial velocity 53\,\text{mph}$at an angle of$60^\circ above horizontal. Find horizontal and vertical velocity component magnitudes.\n - Step 1: Horizontal component magnitude |\mathbf{V}_x|:\n    |\mathbf{V}_x| = 53 \cos(60^\circ) = 53 \times 0.5 = 26.5\,\text{mph} \approx 27\,\text{mph}\n - Step 2: Vertical component magnitude |\mathbf{V}_y|:\n    |\mathbf{V}_y| = 53 \sin(60^\circ) = 53 \times \frac{\sqrt{3}}{2} \approx 53 \times 0.866025 = 45.899\,\text{mph} \approx 46\,\text{mph}\n\n- Vector Magnitude from Component Magnitudes:\n - By the Pythagorean Theorem, total vector magnitude |\mathbf{V}| expressed via components is:\n    |\mathbf{V}| = \sqrt{|\mathbf{V}_x|^2 + |\mathbf{V}_y|^2}\n\n- Example: Arrow Velocity Magnitude and Direction\n - Problem: An arrow is shot into the air with horizontal velocity component 25\,\text{ft/s}$and vertical velocity component$15\,\text{ft/s}. Find the arrow's velocity magnitude and direction angle.\n - Step 1: Calculate magnitude |\mathbf{V}|:\n    |\mathbf{V}| = \sqrt{25^2 + 15^2} = \sqrt{625 + 225} = \sqrt{850} \approx 29.155\,\text{ft/s} \approx 29\,\text{ft/s}\n - Step 2: Calculate direction angle \theta above horizontal:\n    \tan(\theta) = \frac{|\mathbf{V}_y|}{|\mathbf{V}_x|} = \frac{15}{25} = 0.6\n    \theta = \arctan(0.6) \approx 30.96^\circ \approx 31^\circ$$