Problem 18
Question
Given the indicated parts of triangle \(A B C\) with \(\gamma=90^{\circ},\) express the third part in terms of the first two. $$\beta, c ; \quad b$$
Step-by-Step Solution
Verified Answer
\(b = c \cdot \sin(\beta)\)
1Step 1: Understand the Triangle
We have a right triangle with angles \(\alpha\), \(\beta\), and \(\gamma = 90^\circ\) where \(c\) is the hypotenuse, and \(b\) is the side opposite \(\beta\). We need to express side \(b\) in terms of \(\beta\) and \(c\).
2Step 2: Use the Definition of Sine
In a right triangle, the sine of angle \(\beta\) is defined as the ratio of the length of the opposite side to the hypotenuse. This gives us: \( \sin(\beta) = \frac{b}{c} \).
3Step 3: Solve for \(b\)
Rearrange the equation from Step 2 to solve for \(b\). Multiply both sides by \(c\) to isolate \(b\): \( b = c \cdot \sin(\beta) \). This expresses the side \(b\) in terms of \(\beta\) and \(c\).
Key Concepts
Right TriangleSine FunctionHypotenuseOpposite Side
Right Triangle
A right triangle is a special type of triangle where one of the angles is exactly 90 degrees. This 90-degree angle is called a right angle, and it gives the right triangle its name. In a right triangle, the side opposite the right angle plays a unique role, known as the hypotenuse. The other two sides contribute as the legs of the triangle, which are either adjacent or opposite to the other angles in the triangle.
In a right triangle, the sum of the other two angles will always equal 90 degrees, because the total number of degrees in a triangle is always 180. This property makes it easier to determine the angles of a right triangle if one angle is already known. Understanding these basics can help in solving many problems involving trigonometric ratios, like sine, cosine, and tangent.
In a right triangle, the sum of the other two angles will always equal 90 degrees, because the total number of degrees in a triangle is always 180. This property makes it easier to determine the angles of a right triangle if one angle is already known. Understanding these basics can help in solving many problems involving trigonometric ratios, like sine, cosine, and tangent.
Sine Function
The sine function is an essential part of trigonometry and is particularly useful when working with right triangles. It involves the ratio between the length of the opposite side of an angle and the hypotenuse in a right triangle.
By definition, for an angle \(\beta\) in a right triangle, the sine function is represented as \( \sin(\beta) = \frac{\text{opposite}}{\text{hypotenuse}} \). This ratio remains consistent for any given angle of a right triangle, regardless of the triangle's size. When using the sine function, it's possible to determine one side of a right triangle if the other side and the angle are given. This makes sine a powerful tool for solving geometric problems.
By definition, for an angle \(\beta\) in a right triangle, the sine function is represented as \( \sin(\beta) = \frac{\text{opposite}}{\text{hypotenuse}} \). This ratio remains consistent for any given angle of a right triangle, regardless of the triangle's size. When using the sine function, it's possible to determine one side of a right triangle if the other side and the angle are given. This makes sine a powerful tool for solving geometric problems.
Hypotenuse
The hypotenuse is a key component in the study of right triangles. It is defined as the side opposite the right angle, and it is always the longest side in a right triangle. The relationship between the hypotenuse and the other sides can be explored through the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. However, in trigonometry, the hypotenuse is often used in defining sine, cosine, and tangent ratios.
In the case of the sine function, the length of the hypotenuse is essential, because it serves as the denominator in the sine ratio \( \sin(\beta) = \frac{b}{c} \). This indicates the importance of knowing the hypotenuse when calculating trigonometric values for a right triangle.
In the case of the sine function, the length of the hypotenuse is essential, because it serves as the denominator in the sine ratio \( \sin(\beta) = \frac{b}{c} \). This indicates the importance of knowing the hypotenuse when calculating trigonometric values for a right triangle.
Opposite Side
In a right triangle, the opposite side is always defined relative to a specific angle. For an angle \( \beta \), the side opposite is the one that does not form that angle. The opposite side plays a crucial role in defining trigonometric ratios like sine, as it is used as the numerator in the sine function ratio.
Understanding which side is opposite to a given angle helps immensely in solving trigonometric problems. In our exercise, side \( b \) was identified as opposite to angle \( \beta \). By using the formula for the sine, we expressed \( b \) in terms of the hypotenuse \( c \) and angle \( \beta \) with the equation \( b = c \cdot \sin(\beta) \). This clear connection between angle, hypotenuse, and opposite side showcases how powerful trigonometry can be in finding unknown elements of right triangles.
Understanding which side is opposite to a given angle helps immensely in solving trigonometric problems. In our exercise, side \( b \) was identified as opposite to angle \( \beta \). By using the formula for the sine, we expressed \( b \) in terms of the hypotenuse \( c \) and angle \( \beta \) with the equation \( b = c \cdot \sin(\beta) \). This clear connection between angle, hypotenuse, and opposite side showcases how powerful trigonometry can be in finding unknown elements of right triangles.
Other exercises in this chapter
Problem 17
Find the exact value. (a) cse \(240^{\circ}\) (b) \(\csc \left(-330^{\circ}\right)\)
View solution Problem 17
Find the exact values of the trigonometric functions for the acute angle \(\theta\). $$\sin \theta=\frac{3}{5}$$
View solution Problem 18
Find the amplitude, the period, and the phase shift and sketch the graph of the equation. \(y=3 \cos (3 x-\pi)\)
View solution Problem 18
Use a formula for negatives to find the exact value. $$\text { (a) } \sin \left(-\frac{3 \pi}{2}\right) \quad \text { (b) } \cos \left(-225^{\circ}\right) \quad
View solution