Problem 27 - Algebra and Trigonometry

Question

In Exercises 17–32, two sides and an angle (SSA) of a triangle are given. Determine whether the given measurements produce one triangle, two triangles, or no triangle at all. Solve each triangle that results. Round to the nearest tenth and the nearest degree for sides and angles, respectively. $$ a=12, b=16.1, A=37^{\circ} $$

Step-by-Step Solution

Answer

One triangle is possible with sides \(a =12\), \(b = 16.1\), \(c = 33.1\) and angles \(A = 37^\circ\), \(B = 27.9^\circ\), and \(C = 115.1^\circ\).

Step 1: Determine the existence of a triangle

Compare side \(a\) with the length of the side opposite to angle \(A\) (which is \(b \times \sin(A)\)). If \(a > b \sin(A)\), then we can have two triangles. If \(a = b \sin(A)\), we only have one triangle. If \(a < b \sin(A)\), no feasible triangle can be formed. In this case, \(a = 12\), \(b = 16.1\), and \(A = 37^\circ\). Hence, \(b \sin(A) = 16.1 \times \sin(37^\circ) = 9.8\). Since \(a = 12 > 9.8 = b \sin(A)\), two triangles are possible.

Step 2: Calculate the other angles for the two triangles

Use the law of sines to find angle \(B_1\) and \(B_2\). \(B_1 = \arcsin([a \sin(A)] / b) = \arcsin[(12 \times sin(37^\circ)] / 16.1 = 27.9^\circ\). Angle \(B_2 = 180^\circ - B_1 = 152.1^\circ\). For the first triangle, the third angle \(C_1 = 180^\circ - A - B_1 = 115.1^\circ\). For the second triangle, the third angle \(C_2 = 180^\circ - A - B_2 = -9.1^\circ\), which is not feasible in a triangle. Hence, we ignore the second solution.

Step 3: Solve for the remaining side of the triangle

For the first triangle, apply the law of sines again to calculate for side \(c\). The Law of sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle. This gives us \(c = b \sin(C) / \sin(B) = 16.1 \times \sin(115.1^\circ) / \sin(27.9^\circ) = 33.1\).

Key Concepts

Law of Sines

The Law of Sines is a fundamental principle used to solve for unknowns in triangles, particularly when presented with a Side-Side-Angle (SSA) situation, which often does not provide sufficient information for a straightforward solution as is the case with Side-Angle-Side (SAS) or Angle-Side-Angle (ASA) scenarios. The Law of Sines states that the ratio of the length of a side to the sine of its opposite angle is the same for all sides and angles of a given triangle:
\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]
This relation allows for the calculation of missing angles or sides when at least one angle-side opposite pair is known. For instance, knowing side a, side b, and angle A as in our given problem, we can determine angle B by rearranging the law as follows:
\[ B = \arcsin(\frac{a \cdot \sin(A)}{b}) \]
The Law of Sines is not only used to find missing angles but also to check if a solution is possible (as in our exercise). The law helps to establish whether the provided measurements lead to the formation of one or more valid triangles.

Ambiguous Case

SSA is known as the 'Ambiguous Case' for triangle determination because it presents a scenario where several configurations are possible. Considering two sides and an angle that is not enclosed by them, multiple triangles can sometimes be formed.

When we're given side a, side b, and angle A, several situations can occur:

• If a is equal to b multiplied by the sine of A, that is, a = b sin(A), a unique triangle, with angle A at the vertex, is formed.
• If a is less than b sin(A), no triangle can be formed since side a is too short to 'reach' the other side of the triangle when swung like a pendulum from the fixed side b.
• If a is greater than b sin(A) but less than b, two triangles may be possible because side a, when swinging from the hinge of side b, can intersect in two different locations, creating two different possible triangles.
• If a is greater than b, there is one possible triangle since side a will always 'reach' the opposite side.

In our exercise, we determined that two triangles were possible because a was greater than b sin(A). However, one of the proposed solutions led to an impossible angle for a triangle, showing the importance of careful calculation and consideration of all possibilities when dealing with the Ambiguous Case.

Trigonometric Functions

Trigonometric functions are mathematical tools that relate angles to the sides of right triangles, but they are also defined for any angles and extend their applications to various fields such as wave motion, periodic functions, and even in the context of non-right triangles as in the Law of Sines and Law of Cosines.

The sine function (sin), one of the primary trigonometric functions, for example, is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right triangle. In general terms, for a given angle θ, the sine function is the y-coordinate of the point on the unit circle (circle with radius 1) that corresponds to angle θ when the angle’s vertex is placed at the origin.

In the solution to our triangle problem, sine is used to find the necessary angles and the missing side. Not only does understanding of the sine function enable students to solve for unknown variables in triangle problems, but it also helps to establish the relationship between sides and angles that is key to trigonometry and its applications beyond geometry, such as in physics, engineering, and economics where modeling cyclic or wave-like phenomena is essential.