Problem 21 - Precalculus

Question

Solve each triangle. $$\alpha=40^{\circ}, \beta=35^{\circ}, a=6$$

Step-by-Step Solution

Answer

\( \gamma = 105^{\circ}, b \approx 5.44, c \approx 8.92 \)

Step 1: Determine the third angle

In any triangle, the sum of the angles is always 180°. So, we can find the third angle \( \gamma \) using the formula \( \gamma = 180^{\circ} - \alpha - \beta \). Substitute the known values: \( \gamma = 180^{\circ} - 40^{\circ} - 35^{\circ} = 105^{\circ} \).

Step 2: Use the Law of Sines to find side b

The Law of Sines states \( \frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} \). Substitute the known values: \( \frac{6}{\sin(40^{\circ})} = \frac{b}{\sin(35^{\circ})} \). Solving for \( b \), we get \( b = 6 \cdot \frac{\sin(35^{\circ})}{\sin(40^{\circ})} \), which simplifies to approximately \( b \approx 5.44 \).

Step 3: Use the Law of Sines to find side c

Again using the Law of Sines, \( \frac{a}{\sin(\alpha)} = \frac{c}{\sin(\gamma)} \). Substitute in the known values: \( \frac{6}{\sin(40^{\circ})} = \frac{c}{\sin(105^{\circ})} \). Solving for \( c \), we get \( c = 6 \cdot \frac{\sin(105^{\circ})}{\sin(40^{\circ})} \), which simplifies to approximately \( c \approx 8.92 \).

Key Concepts

Angle Sum Property

One of the foundational principles in geometry is the Angle Sum Property of triangles. This property states that the sum of the three interior angles in any triangle is always exactly \(180^{\circ}\). This means, if you know two angles, the third angle can be easily calculated by subtracting the sum of the known angles from \(180^{\circ}\).
For example, if you have a triangle with angles \(\alpha = 40^{\circ}\) and \(\beta = 35^{\circ}\), you can determine the third angle, \(\gamma\), through the formula:

• \(\gamma = 180^{\circ} - \alpha - \beta\)

Substitute the given values:
\[\gamma = 180^{\circ} - 40^{\circ} - 35^{\circ} = 105^{\circ}\].
It's a basic yet powerful tool in solving for unknown components in a triangle.

Law of Sines

The Law of Sines is an essential tool for solving triangles when you know either two angles and one side or two sides and a non-included angle. It relates the ratios of the length of a side of a triangle to the sine of the angle opposite that side. The law is given by:

• \(\frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)}\)

In practice, this allows you to solve for unknown sides if you already know the values of other sides and their opposing angles. For instance, if you know \(a = 6\), \(\alpha = 40^{\circ}\), and \(\beta = 35^{\circ}\), you can find side \(b\) using:
\[b = 6 \cdot \frac{\sin(35^{\circ})}{\sin(40^{\circ})} \approx 5.44\].
Similarly, you can calculate side \(c\) by extending this method:
\[c = 6 \cdot \frac{\sin(105^{\circ})}{\sin(40^{\circ})} \approx 8.92\].
This comprehensive law provides a systematic approach to navigating different configurations of triangle problems.

Triangle sides calculation

When solving triangles, calculating the side lengths is often a crucial step. Using the Law of Sines, calculated in our prior example, you can find sides \(b\) and \(c\) when side \(a\) and two angles are known. This involves simple algebraic manipulation and understanding of trigonometric functions.

First, ensure you've calculated all angles, like \(\gamma\) from the Angle Sum Property. Then, apply the Law of Sines to find unknown sides. If solving for \(b\), use the formula:
\[b = g \cdot \frac{\sin(\beta)}{\sin(\alpha)}\]
Inserting the given and calculated values provides the result \(b \approx 5.44\). Similarly, for \(c\), use:
\[c = a \cdot \frac{\sin(\gamma)}{\sin(\alpha)}\],
solving to get \(c \approx 8.92\).
Ultimately, the strategic use of the Law of Sines not only simplifies the problem-solving process but also solidifies your grasp on solving for triangle dimensions, making it an invaluable concept for geometry enthusiasts.