Problem 42 - Algebra and Trigonometry

Question

Two direction angles of a vector are given. Find the third direction angle, given that it is either obtuse or acute as indicated. (In Exercises 43 and 44, round your answers to the nearest degree.) $$ \beta=\frac{2 \pi}{3}, \quad \gamma=\frac{\pi}{4} ; \quad \alpha \text { is acute } $$

Step-by-Step Solution

Answer

The acute angle \( \alpha \) is 60 degrees.

Step 1: Understanding the Direction Cosine Formula

The direction cosines of a vector are given by \( \cos \alpha \), \( \cos \beta \), and \( \cos \gamma \). For any vector \( \mathbf{v} \), the sum of the squares of its direction cosines equals 1:\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \] This is because they are derived from the normalized vector components.

Step 2: Calculate \( \cos \beta \) and \( \cos \gamma \)

Given \( \beta = \frac{2\pi}{3} \) and \( \gamma = \frac{\pi}{4} \), we need to find \( \cos \beta \) and \( \cos \gamma \):1. \( \cos \beta = \cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2} \)2. \( \cos \gamma = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)

Step 3: Substitute Known Values into the Formula

Substitute \( \cos \beta = -\frac{1}{2} \) and \( \cos \gamma = \frac{\sqrt{2}}{2} \) into the equation:\[ \cos^2 \alpha + \left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = 1 \]This simplifies to:\[ \cos^2 \alpha + \frac{1}{4} + \frac{2}{4} = 1 \]\[ \cos^2 \alpha + \frac{3}{4} = 1 \]

Step 4: Solve for \( \cos^2 \alpha \)

Rearrange the equation to solve for \( \cos^2 \alpha \):\[ \cos^2 \alpha = 1 - \frac{3}{4} \]\[ \cos^2 \alpha = \frac{1}{4} \]

Step 5: Determine \( \cos \alpha \) and \( \alpha \)

Since \( \cos^2 \alpha = \frac{1}{4} \), this implies:\[ \cos \alpha = \pm \frac{1}{2} \]Because \( \alpha \) is acute, we have \( 0 < \alpha < \frac{\pi}{2} \), so \( \cos \alpha \) must be positive:\[ \cos \alpha = \frac{1}{2} \]If \( \cos \alpha = \frac{1}{2} \), then \( \alpha = \frac{\pi}{3} \) or 60 degrees.

Key Concepts

Direction Cosines

When we talk about direction cosines, we're referring to the cosines of the angles that a vector makes with the coordinate axes. These angles help us understand the orientation of the vector in space. Each axis has an angle associated with it:

• \( \alpha \) for the x-axis,
• \( \beta \) for the y-axis,
• \( \gamma \) for the z-axis.

The direction cosines are represented by \( \cos \alpha \), \( \cos \beta \), and \( \cos \gamma \). They are special because they satisfy the equation:\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]This equation holds because they come from a unit vector, which has a magnitude of 1. In practice, this means all three angles work together to fully describe a vector's direction in three-dimensional space.

Trigonometric Angles

Trigonometry allows us to relate the angles of a triangle to the lengths of its sides, offering a perfect toolkit for understanding direction cosines. In 3D vector problems, knowing the trigonometric identities and how to calculate cosines of given angles is crucial.Consider the angles given in an exercise:

• \( \beta = \frac{2\pi}{3} \) results in \( \cos \beta = -\frac{1}{2} \),
• \( \gamma = \frac{\pi}{4} \) gives us \( \cos \gamma = \frac{\sqrt{2}}{2} \).

The use of these trigonometric functions helps solve equations like \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\), where substituting the known cosines allows solving for the unknown cosine.

Acute Angles

Acute angles are crucial in determining the direction cosines since they are angles less than \( \frac{\pi}{2} \, (90^\circ) \).In vector problems, the nature of the angle influences the sign of the cosine value. For acute angles, \( \cos \alpha \) is positive because the angle is smaller than \(90^\circ\).In the exercise, both given angles \( \beta \) and \( \gamma \) are not acute, but \( \alpha \) is specified as acute.When solving for \( \cos \alpha \), we find:\[ \cos^2 \alpha = \frac{1}{4} \]This provides \( \cos \alpha = \pm \frac{1}{2} \).Since \( \alpha \) is acute, we select the positive value:\[ \cos \alpha = \frac{1}{2} \]Thus, \( \alpha = \frac{\pi}{3} \) or \(60^\circ\), ensuring the angle is acute.