Problem 54 - A Graphical Approach to Precalculus with Limits

Question

Solve triangle. There may be two, one, or no such triangle. $$C=22^{\circ} 50^{\prime}, b=159 \text { millimeters, } c=132 \text { millimeters }$$

Step-by-Step Solution

Answer

Two triangles are possible with sides \(a \approx 263.87\) mm for angles \(A = 129^{\circ} 15^{\prime}\) and \(a \approx 27.74\) mm for \(A = 5^{\circ} 5^{\prime}\).

Step 1: Understand the Given Information

We have a triangle with an angle \(C = 22^{\circ} 50^{\prime}\), and two sides: \(b = 159\) mm and \(c = 132\) mm. The goal is to determine all possible angles and the third side \(a\) if the triangle exists.

Step 2: Use the Law of Sines to Solve for Angle B

Using the Law of Sines: \(\frac{b}{\sin B} = \frac{c}{\sin C}\). Plug in the known values: \(\frac{159}{\sin B} = \frac{132}{\sin(22^{\circ} 50^{\prime})}\). Calculate \(\sin(22^{\circ} 50^{\prime}) \approx 0.38880\). Solve for \(\sin B\): \(\sin B = \frac{159 \times 0.38880}{132} \approx 0.46844\). Then, \(B = \sin^{-1}(0.46844) \approx 27^{\circ} 55^{\prime}\). There could be a second possibility for angle B, which is \(152^{\circ} 5^{\prime}\) (\(180^{\circ} - 27^{\circ} 55^{\prime}\)).

Step 3: Determine Possible Angles A for Each Case of B

Case 1: \(B = 27^{\circ} 55^{\prime}\), then \(A = 180^{\circ} - C - B = 180^{\circ} - 22^{\circ} 50^{\prime} - 27^{\circ} 55^{\prime} = 129^{\circ} 15^{\prime}\). Case 2: \(B = 152^{\circ} 5^{\prime}\), then \(A = 180^{\circ} - C - B = 180^{\circ} - 22^{\circ} 50^{\prime} - 152^{\circ} 5^{\prime} = 5^{\circ} 5^{\prime}\).

Step 4: Validate the Two Triangle Solutions

Verify if both cases can form a triangle. For case 1, the angles are \(A = 129^{\circ} 15^{\prime}, B = 27^{\circ} 55^{\prime}, C = 22^{\circ} 50^{\prime}\), which sum to \(180^{\circ}\). For case 2, the angles are \(A = 5^{\circ} 5^{\prime}, B = 152^{\circ} 5^{\prime}, C = 22^{\circ} 50^{\prime}\), which also sum to \(180^{\circ}\). Therefore, two different triangles can exist.

Step 5: Solve for Side a in Each Triangle

Use the Law of Sines again to find side \(a\) for both cases: Case 1: \(\frac{a}{\sin A} = \frac{b}{\sin B}\). Use \(\sin(129^{\circ} 15^{\prime}) \approx 0.77715\) and \(\sin(27^{\circ} 55^{\prime}) \approx 0.46885\), thus \(a = \frac{159 \times 0.77715}{0.46885} \approx 263.87\) mm. Case 2: Use \(\sin(5^{\circ} 5^{\prime}) \approx 0.08979\) and \(\sin(152^{\circ} 5^{\prime}) \approx 0.51504\), and solve \(a = \frac{159 \times 0.08979}{0.51504} \approx 27.74\) mm.

Key Concepts

Law of Sines

The Law of Sines is a fundamental principle in trigonometry that helps solve triangles, especially when you know either two angles and a side (AAS or ASA) or two sides and a non-included angle (SSA). This law states that in any given triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all three sides. Mathematically, it is expressed as:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

Using the provided example: Given angle \(C\) and sides \(b\) and \(c\), we solve for the unknown angles and side using this law. Start by rearranging the formula to solve for \(\sin B\):

• \( \sin B = \frac{b \cdot \sin C}{c} \)

Once you have \(\sin B\), find the angle \(B\) using the inverse sine function. Remember that this approach may yield two solutions for \(B\), as sine has an ambiguous angle property that can provide two potential angles for the same sine value.

Trigonometry

Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. It is crucial in solving triangles, as it offers various functions and laws, such as the Law of Sines and the Law of Cosines, to find unknown measurements in a triangle.
In the context of the provided problem, trigonometry helps break down the task of finding unknown angles and side lengths into manageable steps. Employing the trigonometric sine function, which relates an angle in a right triangle to the ratio between the length of the side opposite the angle and the hypotenuse, is key to calculating these measurements. Consider that the sine function, \(\sin(\theta)\), can be evaluated via a calculator when dealing with non-right triangles.

• It supports the computation of missing angles and sides in non-right-angled triangles.
• Trigonometric identities and equations allow for determining possible values and comparing results.

By understanding these relationships and applying trigonometric principles alongside tools like calculators, solving triangles becomes much more approachable.

Angle Calculation

Calculating angles in a triangle involves knowing specific sums and relationships. In any triangle, the internal angles must sum to exactly 180 degrees. By knowing two angles, it is straightforward to find the third by simple subtraction:

• \( A = 180^\circ - B - C \)

This method confirms the validity of potential triangles by ensuring the angle sum remains consistent. When using the Law of Sines, any calculated sine value informs possible angles:
1. Use the inverse sine function to find the angle from a sine value.
2. Remember the sine law's ambiguity, forming either one or two potential angle values.
3. Assess the possible triangles by checking if the sum satisfies the triangle property of totaling 180 degrees.
Both possible calculations in the problem above, for angle \(A\) and angle \(B\), are verified by ensuring their sums, including angle \(C\), total 180 degrees, guaranteeing that each set forms a valid triangle.