Problem 3 - Precalculus: Functions and Graphs

Question

Find the exact values of the remaining parts of triangle \(A B C\). $$\alpha=60^{\circ}, \quad \beta=45^{\circ}, \quad b=100$$

Step-by-Step Solution

Answer

\(\gamma = 75^{\circ}, a = 50\sqrt{6}, c = 50(\sqrt{6} + \sqrt{2})\).

Step 1: Sum of Angles in a Triangle

In any triangle, the sum of the interior angles is always 180 degrees. This means that \( \alpha + \beta + \gamma = 180^{\circ} \). We can use this to find \( \gamma \).

Step 2: Calculate Angle \( \gamma \)

Using the formula from Step 1, substitute the given angles: \( \gamma = 180^{\circ} - 60^{\circ} - 45^{\circ} = 75^{\circ} \). So, \( \gamma = 75^{\circ} \).

Step 3: Use the Law of Sines to Find Side a

The Law of Sines states \( \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} \). Substitute the known values: \( \frac{a}{\sin 60^{\circ}} = \frac{100}{\sin 45^{\circ}} \).

Step 4: Solve for Side a

Calculate: \( a = \frac{100 \times \sin 60^{\circ}}{\sin 45^{\circ}} = \frac{100 \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} \). Simplify to find \( a = 100 \times \frac{\sqrt{3}}{\sqrt{2}} = 100 \times \frac{\sqrt{3} \times \sqrt{2}}{2} = 100 \times \frac{\sqrt{6}}{2} = 50 \sqrt{6} \).

Step 5: Use the Law of Sines to Find Side c

Again apply the Law of Sines: \( \frac{c}{\sin \gamma} = \frac{b}{\sin \beta} \). Plugging in the values gives \( \frac{c}{\sin 75^{\circ}} = \frac{100}{\sin 45^{\circ}} \).

Step 6: Solve for Side c

Calculate \( c = \frac{100 \times \sin 75^{\circ}}{\sin 45^{\circ}} = \frac{100 \times \sin 75^{\circ}}{\frac{\sqrt{2}}{2}} \). Using known values \( \sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \). Thus, \( c = 100 \times \frac{\sqrt{6} + \sqrt{2}}{2} = 50(\sqrt{6} + \sqrt{2}) \).

Key Concepts

Sum of Angles in a Triangle

In geometry, one of the fundamental properties of a triangle is that the sum of its interior angles is always 180 degrees. This rule applies to all types of triangles, whether they are equilateral, isosceles, or scalene. It's a key starting point for solving many problems in triangle trigonometry.

To find a missing angle in a triangle, knowing the other two angles is crucial. In our exercise, we have two angles given: \( \alpha = 60^{\circ} \) and \( \beta = 45^{\circ} \). Knowing this, we can easily find \( \gamma \), the angle we need. Simply subtract the sum of the known angles from 180 degrees:

• Step 1: \( \alpha + \beta + \gamma = 180^{\circ} \)
• Step 2: \( 60^{\circ} + 45^{\circ} + \gamma = 180^{\circ} \)
• Solve for \( \gamma \): \( \gamma = 180^{\circ} - 60^{\circ} - 45^{\circ} = 75^{\circ} \).

With this simple calculation, the measure of the third angle \( \gamma \) is determined to be 75 degrees.

Law of Sines

The Law of Sines is a powerful tool in trigonometry used to solve for unknown sides or angles in triangles, especially when dealing with non-right triangles. It states that the ratio of the length of a side to the sine of its opposite angle is the same for all sides of the triangle. The formula is given by:

\[ \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} \]
In cases where two angles and one side (AAS, ASA) are known, the Law of Sines is particularly useful. In our case, we know \( \alpha, \beta, \) and side \( b \), and we want to find sides \( a \) and \( c \). Let's see how this works for side \( a \):

• Apply the Law of Sines: \( \frac{a}{\sin 60^{\circ}} = \frac{100}{\sin 45^{\circ}} \)
• Solve the equation for \( a \): \( a = \frac{100 \times \sin 60^{\circ}}{\sin 45^{\circ}} \).

Following a similar method, we can find side \( c \) as well:

• Use \( \frac{c}{\sin 75^{\circ}} = \frac{100}{\sin 45^{\circ}} \)
• Solve for \( c \): \( c = \frac{100 \times \sin 75^{\circ}}{\sin 45^{\circ}} \).

Angle Calculation

Calculating angles in a triangle is often straightforward once you understand the basic concepts. In our original exercise, the goal was to find the third angle \( \gamma \), using the angles \( \alpha \) and \( \beta \) already provided, through the sum of angles rule. However, solving for sides like \( a \) and \( c \) requires careful application of trigonometric identities and the Law of Sines.

For instance, when calculating side \( a \), critical values need to be determined first, such as \( \sin 60^{\circ} \) which equals \( \frac{\sqrt{3}}{2} \) and \( \sin 45^{\circ} \) which equals \( \frac{\sqrt{2}}{2} \). Plugging these values into the Law of Sines equation helps isolate and calculate \( a \).

• Substitute known sine values: \( a = 100 \times \frac{\sin 60^{\circ}}{\sin 45^{\circ}} = 50 \sqrt{6} \)

Similarly, side \( c \) is calculated after determining \( \sin 75^{\circ} \) using its identity:

• \( \sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4} \)
• Resulting in: \( c = 50(\sqrt{6} + \sqrt{2}) \).