Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 9-16: Given the indicated parts of triangle \(A B C\) with \(\gamma=90^{\circ}\), approximate the remaining parts. $$ \beta=71^{\circ} 51^{\prime}, \quad b=240.0 $$

Step-by-Step Solution

Answer

\(\alpha = 18.15^\circ\), \(a \approx 81.7\), \(c \approx 253.2\).

Step 1: Convert Angle to Decimal Degrees

The angle \(\beta\) is given as \(71^{\circ} 51^{\prime}\). Convert the minutes into decimal degrees:\[\beta = 71 + \frac{51}{60} = 71.85^\circ\]

Step 2: Use Right Triangle Trigonometry

In right triangle \(ABC\), we know angle \(\gamma = 90^\circ\). Therefore, angle \(\alpha\) can be found using:\[\alpha = 90^\circ - \beta\]Substitute \(\beta = 71.85^\circ\):\[\alpha = 90^\circ - 71.85^\circ = 18.15^\circ\]

Step 3: Find Side a Using the Sine Function

The side \(a\) can be found using the sine function,\[\sin(\beta) = \frac{b}{c}\]First, rearrange to solve for \(c\):\[c = \frac{b}{\sin(\beta)}\]Calculate using \(b = 240.0\) and \(\beta = 71.85^\circ\):\[c = \frac{240.0}{\sin(71.85^\circ)} \approx 253.2\]

Step 4: Find Side a Using the Cosine Function

Now use the cosine rule to find side \(a\),\[\cos(\beta) = \frac{a}{c}\]Rearrange for \(a\):\[a = c \cdot \cos(\beta)\]Substitute the known values:\[a = 253.2 \cdot \cos(71.85^\circ) \approx 81.7\]

Step 5: Summarize Remaining Parts

The remaining parts of the triangle are as follows: \(\alpha = 18.15^\circ\), \(c \approx 253.2\), and \(a \approx 81.7\).

Key Concepts

Angle Conversion

When dealing with angles in trigonometry, it's common to encounter angles not just in degrees but in a combination of degrees, minutes, and sometimes seconds. This system is similar to how time is measured. In the exercise, the angle \( \beta = 71^\circ 51^\prime \) \ needs to be converted into decimal degrees to simplify calculations.
Conversion is straightforward:

• There are 60 minutes in a degree, just like there are 60 minutes in an hour.
• To convert minutes into a fraction of a degree, divide the minutes by 60.

For example, for \( \beta = 71^\circ 51^\prime \), the calculation is \[ \beta = 71 + \frac{51}{60} = 71.85^\circ \]This conversion allows us to work with a decimal degree, making it easier to use with trigonometric functions and equations. Ensuring accuracy in this step is crucial to obtaining correct results in subsequent calculations.

Trigonometric Functions

Trigonometric functions are essential tools in understanding and solving triangles. In a right triangle like triangle ABC, these functions allow us to relate the angles to the side lengths. The primary functions used here are sine and cosine.

• Sine Function: This function relates the opposite side of an angle to the hypotenuse. The formula is \( \sin(\beta) = \frac{b}{c} \).
• Cosine Function: This function relates the adjacent side of an angle to the hypotenuse. The formula is \( \cos(\beta) = \frac{a}{c} \).

In our exercise, these functions help us find unknown side lengths. Using the sine function, we rearrange to find the hypotenuse \( c \):\[ c = \frac{b}{\sin(\beta)} \]With known \( b = 240.0 \), we calculate \( c \) once \( \beta \) is in decimal degrees. Similarly, to find the adjacent side \( a \), we rearrange the cosine function:\[ a = c \cdot \cos(\beta) \]This demonstrates how trigonometric functions are used to find missing parts of a triangle when some properties are already known.

Triangle Properties

Right triangles have unique properties that set them apart from other triangles. In any right triangle, one angle is always \( 90^\circ \), simplifying some calculations. Here are some key properties:

• Complementary Angles: In a right triangle, the two non-right angles are complementary. This means their sum is always \( 90^\circ \).
• Pythagorean Theorem: Though not directly used in the provided solution, it's essential for right triangles. It states \( a^2 + b^2 = c^2 \), connecting the side lengths.
• Angle Relationships: Knowing one angle allows finding the other. For example, if \( \gamma = 90^\circ \) and \( \beta = 71.85^\circ \), then \( \alpha = 90^\circ - 71.85^\circ = 18.15^\circ \).

Understanding these properties is crucial for solving problems relating to right triangles. They provide the framework that supports various trigonometric calculations, making it easier to derive unknown values from the known parts.