Problem 25 - Precalculus with Limits

Question

Using the Law of Sines. Use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. $$A=110^{\circ}, \quad a=125, \quad b=100$$

Step-by-Step Solution

Answer

For the given values, we found two possible values for angle B, but only the first - 48.59 degrees - results in a possible triangle. For this case, angle C is 21.41 degrees and side c is 46.63 units.

Step 1: Calculate sine of angle B using Law of Sines

Using the Law of Sines, we can express the relationship between angles A and B and sides a and b as sine B = sinA * (b/a). Substituting the known values; sine B = sin(110) * (100/125) = 0.745.

Step 2: Solve for angle B

Use inverse sine to find the angle whose sine is 0.745, this will result in angle B=48.59 degrees. However, sine being a periodic function, it also has another possible solution within the triangle. This can be found by subtracting our solution from 180 degrees, i.e. B' = 180 - 48.59 = 131.41 degrees.

Step 3: Find angle C and side c for first solution

Given that the angles in a triangle sum up to 180, we can calculate angle C = 180 - 110 - 48.59 = 21.41 degrees. Using the Law of Sines to find side c, we have c = a * (sinC/sinA) = 125 * (sin(21.41)/sin(110)) = 46.63.

Step 4: Find angle C and side c for second solution

For the second possible solution, we calculate angle C' = 180 - 110 - 131.41 = -61.41 degrees. Given that the angles in a triangle sum up to 180 degrees, this is not valid. Therefore, our second solution for angle B's does not result in a possible triangle.

Key Concepts

Solving Triangles

Understanding how to solve triangles is a fundamental aspect of trigonometry. When we refer to 'solving a triangle,' we mean finding the missing lengths and angles of the triangle. To do this, we can use various methods, one of which is the Law of Sines. This approach is particularly useful in cases where we have either two angles and one side (AAS or ASA condition) or two sides and a non-included angle (SSA condition).

In the given exercise, we're faced with the latter configuration, and by applying the Law of Sines, we can find the unknown angles and side. It's crucial to pay attention to round the answers to a specified number of decimal places (in this case, two decimal places) for precision. Moreover, it's worth noting that if we apply the SSA condition, we can possibly end up with two different solutions (or none in certain cases) due to the nature of the sine function's periodicity. This ambiguity must be checked for each problem to ensure we account for all possible triangle solutions.

Triangular Periodic Functions

In trigonometry, the sine function is known as a 'periodic function,' which means it repeats its values in regular intervals, known as periods. For the sine function, the period is 360 degrees (or 2π radians), which implies that the sine of an angle and the sine of an angle plus 360 degrees are the same. This periodic nature leads to multiple solutions in certain trigonometry problems.

When we solve for angles using the sine function as in this exercise, we must keep in mind that there can be two different angles in the range of 0 to 180 degrees sharing the same sine value: one acute and one obtuse. However, only one or sometimes neither of these angles will be applicable to a specific triangle scenario. It is sometimes necessary to examine the context of the triangle to determine which angle is indeed part of the solution—this includes consulting the triangle sum theorem which states that the sum of the interior angles must equal 180 degrees.

Inverse Sine

The inverse sine (sin^-1 or arcsin) is essentially the process of finding an angle when its sine value is known. It is the reverse operation of finding the sine of an angle. This function is crucial when solving for unknown angles in a triangle when we have the lengths of the sides.

However, it’s important to understand that the inverse sine function on calculators or in most mathematical software will only provide one of the possible angles—usually the acute one. This is because the range of the principal values for arcsin is typically restricted to between -90 degrees and 90 degrees or between -π/2 and π/2 radians. Hence, when looking for solutions of triangles that could include obtuse angles, we have to remember to explore the possibility of the supplementary angle, i.e., 180 degrees minus the calculated angle, to ensure we are not missing out on potential solutions as demonstrated in the provided exercise.