Geometry, Trigonometry, and Logarithm Applications

This lesson is divided into two distinct components:
- Part A examines the connection between geometry and trigonometry by reviewing concepts from Lessons 12 and 19.
- Part B focuses on using a calculator for logarithms, building upon foundational concepts introduced in Lesson 79.

This section combines knowledge of angle relationships and transversals (Lesson 12) with basic trigonometry (Lesson 19) to solve problems involving right triangles.
There are no new rules or definitions introduced in this section; the focus is on the application of existing mathematical principles to more complex diagram setups.

Parallel lines are indicated in diagrams by arrow tips located on the body of the lines rather than at the ends.
A transversal cutting across parallel lines creates specific angle relationships:
- Adjacent angles created by the transversal are supplementary, meaning they sum to $180^{\circ}$ .
- Inside a system of parallel lines and a transversal, all acute angles are congruent to one another, and all obtuse angles are congruent to one another.

The problem asks to find the values of $x$ and $y$ and round to one decimal place.
Initial Observations:
- An obtuse angle of $120^{\circ}$ is provided outside the right triangle.
- A hypotenuse is identified with a length of $7$ .
- A right triangle is indicated by a square symbol in the corner.
Calculating the Internal Angle:
- Since the angles are supplementary: $180^{\circ} - 120^{\circ} = 60^{\circ}$ .
- Because the lines are parallel, the corresponding acute angle inside the right triangle is also $60^{\circ}$ .
Solving for y (the opposite side):
- Equation: $\sin(60^{\circ}) = \frac{y}{7}$
- Rearranged: $y = 7 \times \sin(60^{\circ})$
- Values: $\sin(60^{\circ}) \approx 0.866$
- Calculation: $7 \times 0.866 = 6.062$
- Rounded Result: $y = 6.1$
Solving for x (the adjacent side):
- Equation: $\cos(60^{\circ}) = \frac{x}{7}$
- Rearranged: $x = 7 \times \cos(60^{\circ})$
- Values: $\cos(60^{\circ}) = 0.5$
- Calculation: $7 \times 0.5 = 3.5$
- Result: $x = 3.5$

The target is to find $x$ and $y$ given a system where parallel lines and a transversal create an angle of $35^{\circ}$ .
Context:
- An angle of $35^{\circ}$ is given at the base of the transversal.
- By the rule of congruent acute angles in transversals, the angle inside the visible right triangle is also $35^{\circ}$ .
- The hypotenuse is given as $20$ .
Finding x (Adjacent Side):
- Equation: $\cos(35^{\circ}) = \frac{x}{20}$
- Calculation: $x = 20 \times \cos(35^{\circ})$
- Rounded Result: $x = 16.4$
Finding y (Opposite Side):
- Equation: $\sin(35^{\circ}) = \frac{y}{20}$
- Calculation: $y = 20 \times \sin(35^{\circ})$
- Rounded Result: $y = 11.5$
Conceptual Check: By visual inspection, $x$ is longer than $y$ in the diagram. Since $16.4 > 11.5$ , the calculated answer is consistent with the scale of the drawing.
Calculator Warning: If calculated results vary significantly (e.g., getting $11.4$ vs $11.5$ due to rounding or very different results), ensure the calculator is set to "degrees" mode rather than "radians."

Logarithms are used to study number properties and simplify calculations involving exceptionally large or small numbers.
Historical Context: John Napier invented logarithms to simplify tedious calculations.

Common Logarithm:
- Notation: $\text{log}$
- Definition: Shorthand for $\text{log}_{10}(n)$ , or the base $10$ logarithm of an argument $n$ .
- Similar to variables where $x$ is understood as $1x$ or $\frac{x}{1}$ , the base $10$ is understood when no base is written.
Natural Logarithm:
- Notation: $\ln$
- Definition: Shorthand for $\text{log}_e(n)$ , where $e$ is the constant approximately equal to $2.718$ .
- The notation "ln" stands for "logarithm natural."

Astronomy (Large Numbers): Measuring distances such as from the Earth to the Sun, which is approximately $93,000,000\,\text{miles}$ .
Chemistry (Small Numbers): Measuring acid concentrations, such as $0.0001\,\text{moles per liter}$ . Logarithms make these numbers more intuitive to work with.
pH Scale: Used in chemistry to determine if a substance is acidic or basic. A neutral pH is $7$ , a value derived from a logarithmic formula.

a) Calculate log 10,000:
- Definition: Finding the exponent needed for $10$ to equal $10,000$ .
- Calculation: $10^4 = 10,000$ .
- Result: $4$
b) Calculate ln 10,000:
- Calculation on Calculator: Type $10,000$ then press the $\ln$ button.
- Full Value: $9.21034$
- Rounded Result: $9.21$
Comparison: The results differ because the bases are different. Because the base of the natural log ( $e \approx 2.718$ ) is smaller than the base of the common log ( $10$ ), it requires a larger exponent ( $9.21$ vs. $4$ ) to reach the same value of $10,000$ .
Verification: Using $e \approx 2.718$ and raising it to the power of $9.21$ yields approximately $9,987$ . The difference of $13$ is due to rounding both the base and the exponent.

Problem: Find the common log and natural log of $93,000,000$ (distance to the Sun).
Common Log:
- Calculation: $\text{log}(93,000,000)$
- Result: $7.97$
Natural Log:
- Calculation: $\ln(93,000,000)$
- Result: $18.35$
Writing $18.35$ is significantly faster and more manageable than writing out $93,000,000$ in scientific and mathematical documentation.

Problem: Find the common log and natural log of $0.0001$ .
Common Log:
- Calculation: $\text{log}(0.0001)$
- Result: $-4$ (Because $10^{-4} = 0.0001$ ).
Natural Log:
- Calculation: $\ln(0.0001)$
- Result: $-9.21$
Working with numbers like $-9.21$ provides more clarity than working with decimal strings like $0.0001$ .

Scientific Calculators: Typically require typing the argument first ( $n$ ), then pressing the $\text{log}$ or $\ln$ button.
Graphing Calculators (e.g., Texas Instruments): Typically require pressing the function button ( $\text{log}$ or $\ln$ ) first, then typing the argument and hitting enter.
On some computer-based calculators (like Macintosh), the button for log base $10$ may explicitly show a script subscript ( $\text{log}_{10}$ ) to distinguish it from other specialized functions like $\text{log}_2$ .