Problem 31 - Algebra and Trigonometry

Question

In triangle \(A B C, \angle A=40^{\circ}, a=15,\) and \(b=20\) (a) Show that there are two triangles, \(A B C\) and \(A^{\prime} B^{\prime} C,\) that satisfy these conditions. (b) Show that the areas of the triangles in part (a) are proportional to the sines of the angles \(C\) and \(C,\) that is, $$\frac{\text { area of } \triangle A B C}{\text { area of } \triangle A^{\prime} B^{\prime} C^{\prime}}=\frac{\sin C}{\sin C^{\prime}}$$

Step-by-Step Solution

Answer

There are two triangles possible; their areas are proportional to \( \sin C \) and \( \sin C' \).

Step 1: Understanding the Ambiguous Case

In a triangle with given angle (here \( \angle A = 40^{\circ} \)), side opposite to the angle (here \( a = 15 \)), and another side (here \( b = 20 \)), this is known as the ambiguous case for the law of sines. When \( a < b \), there can be two solutions or triangles.

Step 2: Apply the Law of Sines to Find Possible Angles

Using the law of sines, first find \( \angle B \):\[ \frac{a}{\sin A} = \frac{b}{\sin B} \] Substituting the known values: \[ \frac{15}{\sin 40^{\circ}} = \frac{20}{\sin B} \] \[ \sin B = \frac{20 \cdot \sin 40^{\circ}}{15} \] Calculate to find \( \sin B \). Since \( \sin B > 1 \) for a certain range, there could be an additional possible angle solution.

Step 3: Calculate the Two Possible Angles for B

Calculate \( \sin B = \frac{20 \cdot \sin 40^{\circ}}{15} \approx 0.855 \).Thus, the possible angles for \( B \) are \( B = \sin^{-1}(0.855) \approx 58.9^{\circ} \) and the second possibility is, \( B^{\prime} = 180^{\circ} - 58.9^{\circ} \approx 121.1^{\circ} \).

Step 4: Determine Angle C for Both Triangles

Since \( A + B + C = 180^{\circ} \), calculate \( C \) for both scenarios:1. \( C \approx 180^{\circ} - 40^{\circ} - 58.9^{\circ} = 81.1^{\circ} \)2. \( C^{\prime} \approx 180^{\circ} - 40^{\circ} - 121.1^{\circ} = 18.9^{\circ} \)

Step 5: Assess Area of Each Triangle

The area of a triangle can be calculated using the formula \( \text{Area} = \frac{1}{2}ab\sin C \). Calculate the area of both triangles:1. For \( \triangle ABC \):\[ \text{Area} = \frac{1}{2} \times 15 \times 20 \times \sin(81.1^{\circ}) \]2. For \( \triangle A'B'C' \):\[ \text{Area'} = \frac{1}{2} \times 15 \times 20 \times \sin(18.9^{\circ}) \]

Step 6: Establish Proportional Relationship of Areas

The relationship is established by the formula:\[ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle A'B'C'} = \frac{\frac{1}{2}ab\sin 81.1^{\circ}}{\frac{1}{2}ab\sin 18.9^{\circ}} = \frac{\sin 81.1^{\circ}}{\sin 18.9^{\circ}} \]This shows the areas are proportional to the sines of angles \( C \) and \( C^{\prime} \).

Key Concepts

Law of Sines

The Law of Sines is a powerful tool in trigonometry, especially when dealing with unknown components of triangles. For a triangle with sides and angles denoted as \( a, b, c \) and \( A, B, C \) respectively, the law states:\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]This relationship enables us to find missing angles or sides when certain elements of a triangle are known.

For instance, in our triangle \( ABC \), given \( \angle A = 40^{\circ} \), \( a = 15 \), and \( b = 20 \), we used the Law of Sines to determine \( \angle B \). By substituting the known values, we get:
- \( \frac{15}{\sin 40^{\circ}} = \frac{20}{\sin B} \)
Solving for \( \sin B \) allows us to find two potential angles due to the ambiguous case, which can occur when an angle and non-included sides are given. This situation allows for two possible triangles, demonstrating the utility yet complexity of the Law of Sines.

Triangle Area Calculation

Calculating the area of a triangle can be done using different methods, but one efficient way is by utilizing sine. For any triangle, the formula using sine is:
- \( \text{Area} = \frac{1}{2}ab\sin C \)
where \( a \) and \( b \) are sides and \( C \) is the included angle. In our example, we apply this formula to both potential triangles \( ABC \) and \( A'B'C' \). The areas are calculated as follows:
- For \( \triangle ABC \): \( \text{Area} = \frac{1}{2} \times 15 \times 20 \times \sin(81.1^{\circ}) \)
- For \( \triangle A'B'C' \): \( \text{Area} = \frac{1}{2} \times 15 \times 20 \times \sin(18.9^{\circ}) \)
This approach highlights how side lengths and angle measures come together to generate the space within a triangle, illustrating a harmony in geometry which the sine function elegantly links.

Angle Proportionality

Angle proportionality in triangles refers to how certain elements, like areas, can correspond directly to other components, such as the sine of an angle. In trigonometry, this proportionality derives from the cosine rule and the Law of Sines.

In our example with triangles \( ABC \) and \( A'B'C' \), the areas exhibit proportionality with the sines of their angles \( C \) and \( C' \). This is shown as:
- \( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle A'B'C'} = \frac{\sin 81.1^{\circ}}{\sin 18.9^{\circ}} \)
Such a relationship suggests that even though two triangles can be quite different in shape, their areas can relate directly to the sine of a particular angle. This concept underscores the intrinsic links within triangles that shape their mathematical beauty and symmetry.