Problem 22
Question
Given the indicated parts of triangle \(A B C\) with \(\gamma=90^{\circ},\) express the third part in terms of the first two. $$\boldsymbol{\beta}, \boldsymbol{a} ; \boldsymbol{c}$$
Step-by-Step Solution
Verified Answer
\( b = a \cdot \cot(\beta) \).
1Step 1: Introduction: Understand Triangle Properties
Triangle \(ABC\) is a right triangle with \(\gamma = 90^\circ\). In a right triangle, we use the Pythagorean theorem and trigonometric identities. For right triangles \(\sin\), \(\cos\), and \(\tan\) of angles are useful: \(\sin(\beta) = \frac{a}{c}\), \(\cos(\beta) = \frac{b}{c}\). Our given parts are angle \(\beta\), side \(a\), and hypotenuse \(c\), and we want to express \(b\) in terms of \(\beta, a,\) and \(c\).
2Step 1: Use Trigonometric Identity for \( \cos \beta \)
Given \(\beta\) and \(c\), use the trigonometric identity for cosine: \[ \cos(\beta) = \frac{b}{c} \]. Solve for \(b\): \[ b = c \cdot \cos(\beta) \].
3Step 2: Express \( c \) in Terms of \( a \) and Trigonometric Functions
Since \(\sin(\beta) = \frac{a}{c}\), rearrange to express \(c\) in terms of \(a\): \[ c = \frac{a}{\sin(\beta)} \].
4Step 3: Combine the Expressions
Substitute \(c\) from Step 2 into the expression found in Step 1: \[ b = \left(\frac{a}{\sin(\beta)}\right) \cdot \cos(\beta) \]. Simplify the expression using the identity \( \cot(\beta) = \frac{\cos(\beta)}{\sin(\beta)}\): \[ b = a \cdot \cot(\beta) \].
Key Concepts
Right TrianglePythagorean TheoremTrigonometric Identities
Right Triangle
A right triangle is a type of triangle that has one of its angles measuring exactly 90 degrees. This angle is known as the right angle. The side opposite this angle is called the hypotenuse, and it is usually the longest side of the triangle. The other two sides are known as the legs of the triangle. A right triangle has some interesting properties:
- The sum of the angles in a triangle is always 180 degrees. Therefore, the other two angles sum up to 90 degrees.
- The longest side, or hypotenuse, is always opposite the right angle.
- Right triangles make trigonometric functions remarkably useful.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle specifically used with right triangles. It relates the lengths of the three sides of a right triangle: If you label the hypotenuse as \(c\) and the other two sides as \(a\) and \(b\), the theorem can be expressed as:\[ a^2 + b^2 = c^2 \]This theorem provides an essential formula for calculating the length of any one side given the lengths of the other two. It's especially useful in many areas of mathematics and physics because:
- It helps find the distance between two points in a plane.
- It supports geometrical constructions and proofs.
- It serves as the basis for trigonometric functions by defining consistent relationships between side lengths.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for any angle. In a right triangle setup, they are particularly powerful tools for relating the angles with the ratio of the triangle's sides:- **Sine** (\(\sin\)) of an angle \(\beta\) is the ratio of the opposite side to the hypotenuse: \(\sin(\beta) = \frac{a}{c}\).
- This allows you to solve for unknown side lengths or angles when only partial information is given.
- Similar to sine, this identity helps determine unknown geometric features of the triangle efficiently.
Other exercises in this chapter
Problem 21
Find the exact values of the trigonometric functions for the acute angle \(\theta\). $$\sec \theta=\frac{6}{5}$$
View solution Problem 22
Verify the identity by transforming the left hand side into the right-hand side. $$\csc (-x) \cos (-x)=-\cot x$$
View solution Problem 22
Find the amplitude, the period, and the phase shift and sketch the graph of the equation. \(y=3 \cos \frac{\pi}{2} x\)
View solution Problem 22
Find the period and sketch the graph of the equation. Show the asymptotes. $$y=\cot \frac{1}{2} x$$
View solution