Problem 19
Question
Exer. 17-24: Given the indicated parts of triangle \(A B C\) with \(\gamma=90^{\circ}\), express the third part in terms of the first two. $$ \beta, b ; \quad a $$
Step-by-Step Solution
Verified Answer
The third part, side \(a\), is expressed as \( a = \frac{b}{\tan(\beta)} \).
1Step 1: Identify the Right Triangle Parts
Given triangle \(ABC\) with \(\gamma = 90^{\circ}\), we have one angle \(\beta\) and the side \(b\) (opposite angle \(\beta\)) as known parts, and we need to find side \(a\) (adjacent to \(\beta\)).
2Step 2: Apply Trigonometric Ratios
In a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side. Therefore, for angle \(\beta\), \( \tan(\beta) = \frac{b}{a} \).
3Step 3: Solve for the Unknown Side
Rearrange the equation to solve for the side \(a\). This gives \( a = \frac{b}{\tan(\beta)} \).
Key Concepts
Right TriangleTrigonometric RatiosAngle Beta
Right Triangle
A right triangle is a special type of triangle in geometry where one of the angles is exactly 90 degrees, known as a right angle. This distinctive feature creates a unique set of side relationships and angle definitions.
- **Vertices and Sides**: The triangle is labeled with vertices A, B, and C, often positioned such that \( \gamma = 90^\circ \) is at vertex C.
- **Hypotenuse**: The side opposite the right angle, often the longest side, is known as the hypotenuse. In triangle \( ABC \), this would commonly be side \( c \).
- **Legs**: The other two sides, adjacent to the right angle, are called the legs. For instance, side \( b \), opposite the angle \( \beta \), and side \( a \), alongside angle \( \beta \), comprise these legs. Right triangles are foundational in trigonometry because they allow for the derivation of several fundamental trigonometric relationships.
- **Vertices and Sides**: The triangle is labeled with vertices A, B, and C, often positioned such that \( \gamma = 90^\circ \) is at vertex C.
- **Hypotenuse**: The side opposite the right angle, often the longest side, is known as the hypotenuse. In triangle \( ABC \), this would commonly be side \( c \).
- **Legs**: The other two sides, adjacent to the right angle, are called the legs. For instance, side \( b \), opposite the angle \( \beta \), and side \( a \), alongside angle \( \beta \), comprise these legs. Right triangles are foundational in trigonometry because they allow for the derivation of several fundamental trigonometric relationships.
Trigonometric Ratios
Trigonometric ratios are core to understanding the relationships between the angles and sides of right triangles. They allow us to express these relationships in a simple mathematical form. Each basic trigonometric ratio corresponds to a particular perspective from an angle in a right triangle.
- **Sine (sin)**: Defined as the ratio of the length of the opposite side to the hypotenuse, \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \).
- **Cosine (cos)**: Expresses the ratio of the adjacent side to the hypotenuse, \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \).
- **Tangent (tan)**: This ratio is the focus of our problem, defining the relationship between the opposite and adjacent sides: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).
These ratios help us solve for unknown sides or angles using known values, making them indispensable tools in trigonometry.
- **Sine (sin)**: Defined as the ratio of the length of the opposite side to the hypotenuse, \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \).
- **Cosine (cos)**: Expresses the ratio of the adjacent side to the hypotenuse, \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \).
- **Tangent (tan)**: This ratio is the focus of our problem, defining the relationship between the opposite and adjacent sides: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).
These ratios help us solve for unknown sides or angles using known values, making them indispensable tools in trigonometry.
Angle Beta
In a right triangle, knowing the properties of one of its non-right angles, such as \( \beta \), can give us profound insights into the triangle's structure. Angle \( \beta \) is typically used as a reference angle for calculating side lengths in right triangles using trigonometric ratios.
- **Relation to Opposite and Adjacent Sides**: In our triangle situation, \( \beta \) is opposite to side \( b \), and adjacent to side \( a \). This association directly impacts the choice of trigonometric ratio applied.
- **Using Tangent Ratio**: When we say \( \tan(\beta) = \frac{b}{a} \), it encapsulates all the information we need. This formula allows us to elucidate unknown side lengths using the measure of \( \beta \) and either of the side lengths \( a \) or \( b \).
- **Practical Calculation**: With existing values for \( \beta \) and \( b \), we solve for \( a \) using \( a = \frac{b}{\tan(\beta)} \). Knowing \( \beta \) enables us to use trigonometric tables or calculators to find \( \tan(\beta) \), solving problems efficiently.
- **Relation to Opposite and Adjacent Sides**: In our triangle situation, \( \beta \) is opposite to side \( b \), and adjacent to side \( a \). This association directly impacts the choice of trigonometric ratio applied.
- **Using Tangent Ratio**: When we say \( \tan(\beta) = \frac{b}{a} \), it encapsulates all the information we need. This formula allows us to elucidate unknown side lengths using the measure of \( \beta \) and either of the side lengths \( a \) or \( b \).
- **Practical Calculation**: With existing values for \( \beta \) and \( b \), we solve for \( a \) using \( a = \frac{b}{\tan(\beta)} \). Knowing \( \beta \) enables us to use trigonometric tables or calculators to find \( \tan(\beta) \), solving problems efficiently.
Other exercises in this chapter
Problem 18
Find the exact values of the trigonometric functions for the acute angle \(\theta\). $$\cos \theta=\frac{8}{17}$$
View solution Problem 18
Express \(\theta\) in terms of degrees, minutes, and seconds, to the nearest second. $$\theta=1.5$$
View solution Problem 19
Exer. 5-40: Find the amplitude, the period, and the phase shift and sketch the graph of the equation. $$ y=\sin \left(\frac{1}{2} x-\frac{\pi}{3}\right) $$
View solution Problem 19
Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=\cot \left(x-\frac{\pi}{2}\right) $$
View solution