Problem 22
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. $$ \beta, a ; \quad c $$
Step-by-Step Solution
Verified Answer
Side \( b \) can be expressed as \( b = \sqrt{c^2 - a^2} \) or \( b = c \cos(\beta) \).
1Step 1: Identify the Known Elements
We are given a right triangle where \( \gamma = 90^{\circ} \). The given parts are the angle \( \beta \), side \( a \), and the hypotenuse \( c \). We need to find the expression for side \( b \) in terms of the given elements.
2Step 2: Use the Pythagorean Theorem
Since triangle \( ABC \) is a right triangle, the Pythagorean theorem states that \( a^2 + b^2 = c^2 \). This means you can solve for \( b \) as follows:\[ b^2 = c^2 - a^2 \]
3Step 3: Solve for Side \( b \)
To find \( b \), take the square root of both sides of the equation derived from the Pythagorean theorem:\[ b = \sqrt{c^2 - a^2} \]
4Step 4: Consider Trigonometric Relations
Additionally, you can use trigonometric relationships to express \( b \) in terms of \( \beta \) and \( c \). The sine relation in a right triangle gives \( \sin(\beta) = \frac{a}{c} \), leading to \( a = c \sin(\beta) \). The cosine relation gives \( \cos(\beta) = \frac{b}{c} \), so \( b = c \cos(\beta) \).
Key Concepts
Pythagorean TheoremTrigonometric RatiosAngle Relationships in Triangles
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry, especially useful when dealing with right triangles. It relates the lengths of the sides in a right triangle and enables us to calculate the length of any side when the lengths of the other two sides are known. The theorem states:
\[ a^2 + b^2 = c^2 \]
where \(c\) is the hypotenuse, the longest side opposite the right angle, and \(a\) and \(b\) are the two legs of the triangle. In the solution provided, this theorem is used to express the unknown side \(b\) in terms of the other known sides, \(a\) and \(c\). This involves rearranging the formula to:
\[ b = \sqrt{c^2 - a^2} \]
This manipulation allows for the calculation of side \(b\) using simple algebraic operations.
\[ a^2 + b^2 = c^2 \]
where \(c\) is the hypotenuse, the longest side opposite the right angle, and \(a\) and \(b\) are the two legs of the triangle. In the solution provided, this theorem is used to express the unknown side \(b\) in terms of the other known sides, \(a\) and \(c\). This involves rearranging the formula to:
\[ b = \sqrt{c^2 - a^2} \]
This manipulation allows for the calculation of side \(b\) using simple algebraic operations.
Trigonometric Ratios
Trigonometric ratios are powerful tools in understanding and working with triangles, especially right triangles. They relate the angles and sides of the triangle, providing different ways to calculate the unknowns. There are three primary trigonometric functions: sine, cosine, and tangent. For a right triangle:
- Sine of an angle \(\beta\) is \(\sin(\beta) = \frac{\text{opposite side}}{\text{hypotenuse}}\)
- Cosine of an angle \(\beta\) is \(\cos(\beta) = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
- Tangent of an angle \(\beta\) is \(\tan(\beta) = \frac{\text{opposite side}}{\text{adjacent side}}\)
Angle Relationships in Triangles
In any triangle, the sum of the interior angles is always \(180^{\circ}\). For right triangles specifically, one of those angles is always \(90^{\circ}\), which simplifies the relationship between the remaining two angles. Therefore, if you know one of the angles, you can easily find the other because:
\[ \alpha + \beta + \gamma = 180^{\circ} \]
Given \(\gamma = 90^{\circ}\), we find that \(\alpha + \beta = 90^{\circ}\). Knowing just one angle in the right triangle automatically gives you the third angle:
\[ \alpha + \beta + \gamma = 180^{\circ} \]
Given \(\gamma = 90^{\circ}\), we find that \(\alpha + \beta = 90^{\circ}\). Knowing just one angle in the right triangle automatically gives you the third angle:
- If you know \(\beta\), you can find \(\alpha = 90^{\circ} - \beta\)
- If you know \(\alpha\), you can find \(\beta = 90^{\circ} - \alpha\)
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 21
Express the angle as a decimal, to the nearest ten-thousandth of a degree. $$37^{\circ} 41^{\prime}$$
View solution Problem 22
Exer. 21-26: Verify the identity by transforming the lefthand side into the right-hand side. $$ \csc (-x) \cos (-x)=-\cot x $$
View solution Problem 22
Exer. 5-40: 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