Problem 31 - Precalculus with Limits

Question

In Exercises \(25-34,\) use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. $$ A=120^{\circ}, \quad a=b=25 $$

Step-by-Step Solution

Answer

The first solution of the triangle is \( A = 120^{\circ} \), \( B = 60^{\circ} \), and \( C = 0^{\circ} \). There is no second solution as in that case, the calculated value for angle \( C \) becomes negative which is not possible in a triangle.

Step 1: Calculate an Angle Using the Law of Sines

Use the Law of Sines to find an angle. As \( a = b \), we can find \( B \) using the formula \( \sin B = \frac{b \cdot \sin A}{a} \). Substitute known values into this equation: \( \sin B = \frac{25 \cdot \sin 120^{\circ}}{25} \). Solve to get \( \sin B \). We get \( \sin B = 0.866 \). This is simplified to \( B = \sin^{-1} (0.866) \) which equals \( 60^{\circ} \).

Step 2: Calculate the Third Angle

Since you have angles \( A \) and \( B \), the third angle (\( C \)) can be calculated by subtracting the sum of \( A \) and \( B \) from \( 180^{\circ} \). This gives us \( C = 180^{\circ} - (120^{\circ} + 60^{\circ}) = 0^{\circ} \).

Step 3: Begin the Second Solution

This problem has mentioned the possibility for a second solution. Calculate \( B \) for this second solution by subtracting the previously calculated \( B\) from \( 180^{\circ} \). This gives \( B = 180^{\circ} - 60^{\circ} = 120^{\circ} \). This is the second possible value for the angle \( B \).

Step 4: Calculate for 'C' in the Second Solution

Now, proceed with the calculations for angle \( C \) using the derived \( B = 120^{\circ} \). You can calculate \( C \) in the same manner as in Step 2. Here, \( C = 180^{\circ} - (120^{\circ} + 120^{\circ}) = -60^{\circ} \). However, a negative angle is not possible for a triangle, hence a second solution is not possible.