Problem 4 - Precalculus: Functions and Graphs

Question

Given the indicated parts of triangle \(A B C\) with \(\gamma=90^{\circ},\) find the exact values of the remaining parts. $$\alpha=60^{\circ}, \quad c=6$$

Step-by-Step Solution

Answer

\(\beta = 30^\circ\), \(a = 3\sqrt{3}\), and \(b = 3\).

Step 1: Understanding the Given Values

We are provided with triangle \(ABC\), which has a right angle at \(\gamma = 90^\circ\). The angle \(\alpha = 60^\circ\) and the side \(c = 6\). We need to find the exact values of the remaining parts, including side \(b\), side \(a\), and the angle \(\beta\).

Step 2: Finding Angle β

Since the sum of angles in a triangle is \(180^\circ\), and \(\gamma = 90^\circ\), we can find angle \(\beta\) using \(\alpha + \beta + \gamma = 180^\circ\). Thus, \(\beta = 180^\circ - \alpha - \gamma = 180^\circ - 60^\circ - 90^\circ = 30^\circ\).

Step 3: Finding Side a using Sine

Using the definition of sine in a right triangle, \( \sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} \). For \(\triangle ABC\), this means \( \sin(60^\circ) = \frac{a}{c} = \frac{a}{6}\). Solving for \(a\), we have \(a = 6 \cdot \sin(60^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3}\).

Step 4: Finding Side b using Cosine

Using the definition of cosine in a right triangle, \( \cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} \). For \(\triangle ABC\), this means \( \cos(60^\circ) = \frac{b}{c} = \frac{b}{6}\). Solving for \(b\), we have \(b = 6 \cdot \cos(60^\circ) = 6 \cdot \frac{1}{2} = 3\).

Key Concepts

Angle Sum Property

In every triangle, the sum of all interior angles is always a constant value of 180 degrees. This rule assists in determining unknown angles when at least two angles are known.
In our given right triangle \(ABC\), we know that angle \( \gamma \) is 90 degrees, as it is a right triangle. We are also given \( \alpha = 60^{\circ} \). Using the angle sum property, we can determine angle \( \beta \):

• Sum of angles = \( \alpha + \beta + \gamma = 180^{\circ} \)
• Substitute the known values into the equation: \( 60^{\circ} + \beta + 90^{\circ} = 180^{\circ} \)
• Solve for \( \beta \) which results in \( \beta = 30^{\circ} \)

This property is essential in trigonometry and is frequently used to solve problems in triangles given limited information.

Sine Function

The sine function connects an angle in a right triangle with the ratio of the length of the side opposite the angle to the hypotenuse. The formula is expressed as: \[\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\]
In \(\triangle ABC\), the sine of \(\alpha\), which is \(60^{\circ}\), is used to find side \(a\):

• Given, \(\sin(60^{\circ}) = \frac{a}{c} = \frac{a}{6}\)
• Account for the exact value of \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\)
• Solve for \(a\): \(a = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\)

The sine function is highly useful in revealing the relationship between triangle angles and their respective sides' lengths. Knowing exact trigonometric values like \(\sin(60^{\circ})\) simplifies calculations.

Cosine Function

The cosine function in trigonometry relates an angle in a right triangle to the ratio of the length of the side adjacent to the angle and the hypotenuse. The formula is given by \[\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]
Applying this formula to \(\triangle ABC\) allows us to find side \(b\) when \(\alpha = 60^{\circ}\):

• \(\cos(60^{\circ}) = \frac{b}{c} = \frac{b}{6}\)
• Use the exact value of \(\cos(60^{\circ}) = \frac{1}{2}\)
• Solve for \(b\), resulting in: \(b = 6 \times \frac{1}{2} = 3\)

The cosine function, similar to the sine function, is vital for understanding side lengths and angle relationships within right-angle triangles.

Exact Values in Trigonometry

In trigonometry, knowing exact values for certain standard angles is crucial for solving right triangles efficiently. These exact values are based on special triangles, such as the 30-60-90 triangle and the 45-45-90 triangle.
Focusing on the angles in our triangle:

• For \(60^{\circ}\): \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\) and \(\cos(60^{\circ}) = \frac{1}{2}\)
• For \(30^{\circ}\), derived from solving for \(\beta\): \(\sin(30^{\circ}) = \frac{1}{2}\) and \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\)

Utilizing these exact trigonometric values simplifies calculations. This knowledge makes determining unknown sides or angles straightforward when dealing with standard angles. Exact values highlight the elegance and simplicity underlying complex trigonometric problems.